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k a July Highlights and 2025 AoPS Online Class Information
jwelsh   0
Jul 1, 2025
We are halfway through summer, so be sure to carve out some time to keep your skills sharp and explore challenging topics at AoPS Online and our AoPS Academies (including the Virtual Campus)!

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0 replies
jwelsh
Jul 1, 2025
0 replies
geometry is mAJorly BACk
EpicBird08   15
N 8 minutes ago by timon92
Source: ISL 2024/G6
Let $ABC$ be an acute triangle with $AB < AC$, and let $\Gamma$ be the circumcircle of $ABC$. Points $X$ and $Y$ lie on $\Gamma$ so that $XY$ and $BC$ meet on the external angle bisector of $\angle BAC$. Suppose that the tangents to $\Gamma$ at $x$ and $Y$ intersect at a point $T$ on the same side of $BC$ as $A$, and that $TX$ and $TY$ intersect $BC$ at $U$ and $V$, respectively. Let $J$ be the centre of the excircle of triangle $TUV$ opposite the vertex $T$.

Prove that $AJ$ bisects $\angle BAC$.
15 replies
1 viewing
EpicBird08
Jul 16, 2025
timon92
8 minutes ago
Elegant Geometry
Euler_Gauss   1
N 11 minutes ago by ezpotd
Let \( O \), \( H \) be the circumcenter and orthocenter of \(\triangle ABC\) respectively, and let \( L \) be the reflection of \( O \) over \( AB \). \( E \) lies on \( AB \), and a line parallel to \( BH \) through \( E \) intersects \(\odot( AEL)\) at \( K \).

Suppose \( KH \) meets \( BC \) at \( D \). Prove that $\angle KLE = \angle KDE.$
1 reply
Euler_Gauss
6 hours ago
ezpotd
11 minutes ago
2025 IMO Results
ilikemath247365   4
N 22 minutes ago by Aiden-1089
Source: https://www.imo-official.org/year_info.aspx?year=2025
Congrats to China for getting 1st place! Congrats to USA for getting 2nd and congrats to South Korea for getting 3rd!
4 replies
ilikemath247365
Yesterday at 4:48 PM
Aiden-1089
22 minutes ago
Inequality
SunnyEvan   4
N 23 minutes ago by SunnyEvan
Source: Own
Let $ a,b,c >0$, such that: $ a+b+c=3 .$ Prove that:
$$ 2025 \geq (628-96(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}))(ab+bc+ca)+864\frac{a^3+b^3+c^3}{a^2+b^2+c^2}$$When does the equality hold ?
4 replies
SunnyEvan
Jul 18, 2025
SunnyEvan
23 minutes ago
Nothing but a game
AlexCenteno2007   1
N 4 hours ago by AlexCenteno2007
Let ABCD be a trapeze with AD ∥ BC. M and N are the midpoints of CD and BC
respectively, and P is the common point of the lines AM and DN. If PM/AP = 4, show that
ABCD is a parallelogram.
1 reply
AlexCenteno2007
Yesterday at 5:11 PM
AlexCenteno2007
4 hours ago
Inequalities
sqing   13
N 5 hours ago by sqing
Let $ a,b,c,ab+bc+ca = 3. $Prove that$$\sqrt[3]{ \frac{1}{a} + 7b} + \sqrt[3]{\frac{1}{b} + 7c} + \sqrt[3]{\frac{1}{c} + 7a } \leq \frac{6}{abc}$$
13 replies
sqing
Jun 18, 2025
sqing
5 hours ago
Inequalities
sqing   18
N 5 hours ago by sqing
Let $ a, b, c $ be real numbers such that $ a + b + c = 0 $ and $ abc = -16 $. Prove that$$ \frac{a^2 + b^2}{c} + \frac{b^2 + c^2}{a} + \frac{c^2 + a^2}{b} -a^2\geq 2$$$$ \frac{a^2 + b^2}{c} + \frac{b^2 + c^2}{a} + \frac{c^2 + a^2}{b}-a^2-bc\geq -2$$Equality holds when $a=-4,b=c=2.$
18 replies
sqing
Jul 9, 2025
sqing
5 hours ago
Play a lot
AlexCenteno2007   2
N Yesterday at 10:56 PM by ohiorizzler1434
In a triangle ABC, let D be the foot of the altitude from A, and M the midpoint of BC. Through M
draw straight lines parallel to AB and AC, which intersect AD at P and Q respectively. Show
that A and D are harmonic conjugates of P and Q.
2 replies
AlexCenteno2007
Tuesday at 6:32 PM
ohiorizzler1434
Yesterday at 10:56 PM
Nice Geometry Proof Question
MathRook7817   5
N Yesterday at 10:49 PM by ohiorizzler1434
Let $ABC$ be an acute triangle with orthocenter $H$. Prove that the triangle formed by the perpendicular bisectors of $AH$, $BH$, and $CH$ is congruent to triangle $ABC$.
5 replies
MathRook7817
Yesterday at 4:32 PM
ohiorizzler1434
Yesterday at 10:49 PM
Group theory study and review
JerryZYang   6
N Yesterday at 8:36 PM by JerryZYang
I want to review and potentially teach group theory... So can anyone give me some nice resources for review and beginners. The review ones are for me with around Middle School reading level the beginner ones are for me to see how to teach. :) thanks!
6 replies
JerryZYang
Yesterday at 4:14 AM
JerryZYang
Yesterday at 8:36 PM
Geometry / trigonometry
Vice8427   1
N Yesterday at 8:28 PM by vanstraelen
You have a triangle ABC, D and E are points in AC so that AD = DE = 2, and EC = 3, now with that points the following triangles are formed, triangle BDC and triangle BEC.
Now a, b and c are the angles on the opposite vertex of side BC on all three triangles (BAC, BDC and BEC) respectively (a is at vertex A, b at vertex D and c at vertex E).
Given that a+b+c= 90° or π\2
What is the lenght of BC?
The question is supposed to be doable with at most basic trigonometry. Not requiring more advanced or complex contents.
Sorry if I made any sort of grammar mistake or couldnt explain myself correctly, I'm a native spanish speaker.

Also, you can't upload images? I couldnt find a button or tool to upload the image of the problem.
1 reply
Vice8427
Jun 9, 2025
vanstraelen
Yesterday at 8:28 PM
Inequality
Ecrin_eren   1
N Yesterday at 6:45 PM by mathprodigy2011


For positive real numbers a, b, c, prove that

[(a + 2b + 2c)(3a + 2b + 2c)(b + c)(2a + b + c)]
+
[(b + 2c + 2a)(3b + 2c + 2a)(c + a)(2b + c + a)]
+
[(c + 2a + 2b)(3c + 2a + 2b)(a + b)(2c + a + b)]
≥ 105/8




1 reply
Ecrin_eren
Yesterday at 2:13 PM
mathprodigy2011
Yesterday at 6:45 PM
Solution set of 2/x>3/3-x
EthanWYX2009   4
N Yesterday at 6:43 PM by mathprodigy2011
The solution set of the inequality \( \frac{2}{x} > \frac{3}{3-x} \) is ______.

Proposed by Baihao Lan, High School Attached to Northwest Normal University
4 replies
EthanWYX2009
Jul 22, 2025
mathprodigy2011
Yesterday at 6:43 PM
Inequality
Ecrin_eren   1
N Yesterday at 4:39 PM by Ecrin_eren
For positive real numbers a, b, c satisfying
ab + ac + bc = 3abc, prove that

bc / (a⁴(b + c)) + ac / (b⁴(a + c)) + ab / (c⁴(a + b)) ≥ 3/2
1 reply
Ecrin_eren
Yesterday at 2:16 PM
Ecrin_eren
Yesterday at 4:39 PM
Weird expression being integer.
MarkBcc168   25
N Jul 11, 2025 by Ilikeminecraft
Source: IMO Shortlist 2017 N5
Find all pairs $(p,q)$ of prime numbers which $p>q$ and
$$\frac{(p+q)^{p+q}(p-q)^{p-q}-1}{(p+q)^{p-q}(p-q)^{p+q}-1}$$is an integer.
25 replies
MarkBcc168
Jul 10, 2018
Ilikeminecraft
Jul 11, 2025
Weird expression being integer.
G H J
G H BBookmark kLocked kLocked NReply
Source: IMO Shortlist 2017 N5
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MarkBcc168
1631 posts
#1 • 6 Y
Y by Amir Hossein, yayitsme, megarnie, Adventure10, Mango247, deplasmanyollari
Find all pairs $(p,q)$ of prime numbers which $p>q$ and
$$\frac{(p+q)^{p+q}(p-q)^{p-q}-1}{(p+q)^{p-q}(p-q)^{p+q}-1}$$is an integer.
This post has been edited 1 time. Last edited by MarkBcc168, Jul 10, 2018, 11:27 AM
Reason: use ISL wording
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MarkBcc168
1631 posts
#2 • 9 Y
Y by Amir Hossein, me9hanics, MatBoy-123, kawazmlekiem, megarnie, ILOVEMYFAMILY, DrYouKnowWho, Adventure10, Mango247
Really straightforward for N5.

Solution
This post has been edited 1 time. Last edited by MarkBcc168, Dec 30, 2020, 2:37 PM
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v_Enhance
6906 posts
#3 • 6 Y
Y by samoha, Amir Hossein, me9hanics, v4913, HamstPan38825, Adventure10
For brevity let $a = p+q$ and $b = p-q$, so $a > b$ and $\gcd(a,b) \le 2$. Then we have in the usual way that \begin{align*} 	a^b b^a - 1 &\mid a^a b^b - 1 \\ 	\implies a^b b^a - 1 &\mid a^a b^b - a^b b ^a \\ 	&= (ab)^b \left( a^{a-b} - b^{a-b} \right) \\ 	\implies a^b b^a - 1 &\mid a^{a-b} - b^{a-b} \\ 	&= a^{2q} - b^{2q} \\ 	&= 4pq \cdot \frac{a^{2q}-b^{2q}}{a^2-b^2}. \end{align*}
Claim: There are no solutions for $q \ge 5$.

Proof. Assume $q > 2$. In that case $a$ and $b$ are even. Let $r = \sqrt{a^b b^a}$, meaning $r$ is even and $r^2-1$ divides the expression above. First, \begin{align*} 	r &= (p+q)^{\frac{1}{2}(p-q)} (p-q)^{\frac{1}{2}(p+q)} \\ 	\implies r &\equiv (-1)^{\frac{1}{2}(p+q)} q^p \pmod p \\ 	&\equiv (-1)^{\frac{1}{2}(p+q)} q \pmod p \end{align*}and since $p \ge q+2$, it follows $r \pm 1 \not\equiv 0 \pmod p$.

Consequently $(r-1)(r+1)$ divides $T = q \cdot \frac{a^{2q}-b^{2q}}{a^2-b^2}$. However by cyclotomic arguments, all prime factors of $T$ are $1 \pmod q$ or $q$.

So we conclude $r-1$ and $r+1$ are in $\{0,1\} \pmod q$. Hence $q \le 3$ as desired. $\blacksquare$

Now we grind through the $q=2$ and $q=3$ cases.
  • For $q=2$, we arrive at \[ a^b b^a - 1 \mid 4(a+b) (a^2+b^2). \]If $p \ge 5$ then $a \ge 7$, $b \ge 3$ and so $a^b b^a \ge a^3 \cdot b^3 \cdot 3^4 > 16a^3b^3$ which is a contradiction. On the other hand $(p,q) = (3,2) \iff (a,b)=(5,1)$ is evidently a solution.
  • Similarly, for $q=3$ we get \[ a^b b^a - 1 \mid 6(a+b)(a^4+a^2b^2+b^4) \]If $p \ge 11$ then $a \ge 14$, $b \ge 8$ and $a^b b^a \ge a^7 b^7 \cdot 8^6 > 36 a^7 b^7$ which is a contradiction. On the other hand neither $(p,q)=(5,3) \iff (a,b)=(8,2)$ nor $(p,q)=(7,3) \iff (a,b) = (10,4)$ are solutions, since:
    • If $(a,b)=(8,2)$ then $a^b b^a = 2^{14}-1$, which is divisible by $43 \mid 2^7+1$. But $6(a+b)(a^4+a^2b^2+b^4) = 6 \cdot 10 \cdot 4368$ is not divisible by $43$.
    • If $(a,b)=(10,4)$ then $a^b b^a = 10^4 \cdot 4^{10} - 1 = 320^4 - 1$, which is divisible by $11$ (dividing $319$). But $6(a+b)(a^4+a^2b^2+b^4) = 6 \cdot 14 \cdot 11856$, which is not divisible by $11$.

In conclusion, the only solution is $(p,q)=(3,2)$ which visibly works.
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anantmudgal09
1980 posts
#4 • 8 Y
Y by Wizard_32, p_square, Pluto1708, khina, Ankoganit, Adventure10, Mango247, math_comb01
Quotes on this at PDC:
"Remember kids, $n^2-1$ factors as $(n-1)(n+1)$."
"Primes are odd". "Uh, except $2$ right?" "No one cares [insert person name]".
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juckter
329 posts
#5 • 2 Y
Y by Adventure10, Mango247
Kind of boring, but ok as N5

Let $M = (p + q)^{p - q}(p - q)^{p + q} - 1$, then $M$ divides $(p + q)^{p + q}(p - q)^{p + q} - (p + q)^2q$. If the problem condition holds then it also divides $(p + q)^{p + q}(p - q)^{p + q} - (p - q)^2q$. Subtracting we find that

$$M \mid (p + q)^{2q} - (p - q)^{2q}$$
Let $r$ be any prime dividing $M$. Then $r \mid (p + q)^{2q} - (p - q)^{2q}$ and $(p + q)^{2q} \equiv (p - q)^{2q} \pmod{r}$. Clearly $r$ is relatively prime to both $p + q$ and $p - q$ because it divides $M$. Then the previous congruence implies that

$$\left(\frac{p + q}{p - q}\right)^{2q} \equiv 1 \pmod{r}$$
Implying that the order of $\frac{p + q}{p - q}$ divides $2q$. We assume for the moment that $q > 2$ and deal with this case at the end. Assume first that the order is not divisible by $q$. Then it divides $2$ and therefore

$$(p + q)^2 \equiv (p - q)^2 \pmod{q} \implies r \mid 4pq$$
This implies that $r = 2$, $r = p$ or $r = q$. The first case is impossible because $q > 2$ implies that $p - q$ and $p + q$ are even and $M$ is odd. Now, if $r = p$ then $p \mid M$ implies that

$$0 \equiv (p + q)^{p - q}(p - q)^{p + q} - 1 \equiv q^{2p} \pmod{p}$$
Implying that the order of $q \pmod{p}$ divides $2p$. Because it is at most $p - 1$ it follows that the order divides $2$ and $p \mid q^2 - 1 = (q - 1)(q + 1)$ which is impossible as $p \geq q + 2$.

Finally if $r = q$ then similarly we find that $p^{2p} \equiv 1 \pmod{q}$ and the order of $p \pmod{q}$ must divide $2$, so $q \mid (p - 1)(p + 1)$ and $p = kq \pm 1$ for some positive integer $k$ and some choice of sign. Because both $p$ and $q$ are odd, it follows that $k$ is even. Now recall that $M$ must divide $(p + q)^{2q} - (p - q)^{2q}$. If $k \geq 4$ then the exponent of $p + q$ in $M$ is greater than $2q$, and $M > (p + q)^{2q} - (p - q)^{2q}$, contradicting that $M$ must divide this expression. Thus $k \leq 3$ and the only possibility is that $k = 2$. We now deal with these two cases.

If $p = 2q + 1$.

Then $M = (q + 1)^{3q + 1}(3q + 1)^{q + 1} - 1$ must divide $(3q + 1)^{2q} - (q + 1)^{2q}$, which implies that $(q + 1)^{3q + 1} \leq (3q + 1)^{q - 1}$. However this is false for every $q$ because

$$(q + 1)^{3q} = ((q + 1)^2(q + 1))^q > (3(q + 1))^q > (3q + 1)^q > (3q + 1)^{q - 1}$$
If $p = 2q - 1$.

Then $M = (q - 1)^{3q - 1}(3q + 1)^{q - 1} - 1$ divides $(3q - 1)^{2q} - (q - 1)^{2q}$. We must now have $(q - 1)^{3q - 1} \leq (3q - 1)^{q + 1}$, so we must have $(q - 1)^{3 - 1/q} \leq (3q - 1)^{1 + 1/q} \leq (4q - 4)^{1 + 1/q}$. The inequality now reduces to

$$(q - 1)^{2 - 2/q} \leq 4^{1 + 1/q}$$
If $q \geq 5$ then $q - 1 \geq 4$ and we must have $2 - 2/q \leq 1 + 1/q$ which gives $q \leq 3$, a contradiction. If $q = 3$ then $p = 5$ which we easiy can prove doesn't satisfy the condition.

In all remaining cases the order of $\frac{p + q}{p - q} \pmod{r}$ must be divisible by $q$, implying that $q$ divides $r - 1$, and therefore every prime divisor of $M$ is congruent to $1$ modulo $q$. Now, we have

$$M = \left((p + q)^{\frac{p - q}{2}}(p - q)^{\frac{p + q}{2}} - 1\right)\left((p + q)^{\frac{p - q}{2}}(p -q)^{\frac{p + q}{2}} + 1\right)$$
Each of these factors must be congruent to $1 \mod q$ because all of their prime divisors is, but this is impossible because they differ by $2$ and $q$ is not $2$.

Finally, if $q = 2$ then $M = (p + 2)^{p - 2}(p - 2)^{p + 2} - 1$ must divide $(p + 2)^4 - (p - 2)^4$. This is impossible for size reasons if $p \geq 7$. If $p = 5$ then $M = 7^33^7 - 1$ divides $7^4 - 3^7$ which once again doesn't hold for size reasons. Finally if $p = 3$ we find that

$$\frac{5^51^1 - 1}{5^11^5 - 1} = \frac{3124}{4} = 781$$
Is an integer, so this is the only pair that works.
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DrMath
2130 posts
#6 • 3 Y
Y by AlgebraFC, Adventure10, Mango247
Subtracting one from the fraction, we get $(p+q)^{p-q}(p-q)^{p+q}-1\mid [(p-q)(p+q)]^{p-q}[(p+q)^{2q}-(p-q)^{2q}]$. Now, if $r$ is a prime and divides $(p+q)(p-q)$, it cannot divide the LHS, so we get $$(p+q)^{p-q}(p-q)^{p+1}-1\mid (p+q)^{2q}-(p-q)^{2q}$$Thus, as a preliminary bound we get $p-q<2q$ or $p<3q$. From here on, assume $q>2$, so both $p,q$ are odd.

Suppose $r$ is a prime that divides the LHS. Then $r$ divides the RHS and is relatively prime to $(p+q)(p-q)$, so we get $r\mid \left(\frac{p+q}{p-q}\right)^{2q}-1$. Now, if $q\nmid r-1$, we get $r\mid (p+q)^2-(p-q)^2=4pq$, so $r\in \{2,p,q\}$. But $2$ cannot divide the LHS, so we are limited to $r=p$ or $r=q$. Suppose $r=p$; then $p\mid (p+q)^{p-q}(p-q)^{p+q}-1$ or $p\mid q^{2p}-1$. Thus $p\mid q^2-1=(q-1)(q+1)$, but this contradicts $p>q$. Thus, the only possibility is $r=q$. Otherwise, $r\equiv 1\pmod{q}$. Thus, every prime factor of the LHS is either $q$ or $1\pmod{q}$.

On the other hand, if we write $X=(p+q)^{\frac{p-q}{2}}(p-q)^{\frac{p+q}{2}}$, then the LHS can be written as $X^2-1$. On the other hand, every factor of $X^2-1$ is either $0$ or $1\pmod{q}$. Thus, $X-1, X+1\in \{0,1\}\pmod{q}$. The only way this can happen for $q>2$ is if $q=3$ and $X\equiv 2\pmod{3}$. Now, since $q=3$ we merely need to test $p<9$, or $p=5,7$. It is easy to check these cases fail. This exhausts all the cases for when $q$ is odd.

Finally, checking $q=2$, we need $p<6$ or $p=3,5$. The only value of $p$ which works is $3$, and we get the ordered pair $\boxed{(3,2)}$.
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pavel kozlov
618 posts
#7 • 3 Y
Y by brokendiamond, Adventure10, Mango247
Let $A=(p+q)^{p+q}(p-q)^{p-q}-1, B=(p+q)^{p-q}(p-q)^{p+q}-1$.
First we get that $q<p\leq3q$ as in the beginning of the first solution and also we do $q=2$ case work.
Secondly we look at any prime divisor $t$ of $B$ and do some casework (excluding cases $t|(p+q)^2-(p-q)^2$ and $t|((p^2-q^2)^2-1$), after what we get that $t \equiv 1 \mod pq$.

Here one can get stucked if he don't notice a simple fact that $B$ if of the form $n^2-1$. But there is a brute force finish in this case. Let's look.

Every prime of $B$ has remainder $1$ modulo $pq$, so $B$ itself has remainder $1$ modulo $pq$.
$B \equiv 1 \mod p$ means $p|q^{2p}-2$ or $p|q^2-2$, so $q^2-2=sp$ with an integer $s$.
Futher, $B \equiv 1 \mod q$ means $p^{2p}\equiv 2 \mod q$, hence $p^{-2}\equiv p^{2(q^2-2)}=p^{2sp}\equiv 2^s \mod q$, or, equivalently, $p^2\equiv 2^{-s} \mod q$.
Taking each part of congruence into degree $p$ we get $-2^{sp}\equiv p^{2p} \equiv 2 \mod q$, so $q|2^{sp-1}+1$.
If $r$ is order of $2$ modulo $q$ then $r|(2(q^2-3),(q-1))\leq 4$, so the only thing left to do is to check primes $q|15$.
This post has been edited 2 times. Last edited by pavel kozlov, Jul 17, 2018, 2:50 PM
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pad
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#8 • 1 Y
Y by Mango247
Suppose $q\not = 2$ for now. Let $a=p+q,b=p-q$. Then $a^bb^a-1\mid a^ab^b-1$, so $a^bb^a-1\mid a^ab^b-a^bb^a = a^bb^b(a^{a-b}-b^{a-b})$. But $\gcd(a^bb^a-1,a^bb^b)=1$, so $a^bb^a-1\mid a^{2q}-b^{2q}$. Suppose a prime $r$ divides $a^bb^a-1$. Note that $r$ is odd since $a,b$ are even. Then $r\mid a^{2q}-b^{2q}$, so the order of $a/b$ mod $r$ divides $2q$, so it is in the set $\{1,2,q,2q\}$.

If the order is 1 or 2, then $a^2\equiv b^2 \pmod{r}$, i.e. $(p+q)^2-(p-q)^2=4pq \equiv 0 \pmod{r}$, so $r \in \{p,q\}$. However, if $r=p$, then
\[ 0\equiv (p+q)^{p-q}(p-q)^{p+q}-1\equiv q^{p-q}\cdot (-q)^{p+q}-1 \equiv q^{2p}-1 \equiv q^2-1 \pmod{p} \]which is not possible, since $p>q$. If $r=q$, then
\[ 0\equiv (p+q)^{p-q}(p-q)^{p+q}-1 \equiv p^{p-q}\cdot p^{p+q}-1 \equiv p^{2p}-1\pmod{q}.\]Now the order of $p \pmod{q}$ divides $2p$, so it is in the set $\{1,2,p,2p\}$. It also divides $q-1$, which is a contradiction if the order is $p$ or $2p$, which means the order of $p\pmod q$ is 1 or 2. But then $p^2\equiv 1 \pmod{q}$, so $q\mid (p-1)(p+1)$, contradiction.

Therefore, the order of $a/b$ mod $r$ is $q$ or $2q$. But it also divides $r-1$, so $q\mid r-1$. Hence, $r\equiv 1 \pmod{q}$, so every prime factor of $a^bb^a-1$ is $q$ or 1 mod $q$. But $a,b$ are even, so $a^bb^a=k^2$ for some $k$. So $a^bb^a-1=(k-1)(k+1)$, which means every prime factor of $k-1,k+1$ are 0,1 mod $q$, and in particular $k-1,k+1$ are both 0 or 1 mod $q$. This is impossible unless $q=2$ or $q=3$.

We conclude that $q \in \{2,3\}$. A check on these both gives that the only solution is $(p,q)=(3,2)$.

Remarks
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TheUltimate123
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#9
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Solved with nukelauncher. Since the denominator is relatively prime with \(p+q\) and \(p-q\), it is equivalent to note that \[\frac1{(p+q)^{p-q}(p-q)^{p-q}}\left(\frac{(p+q)^{p+q}(p-q)^{p-q}-1}{(p+q)^{p-q}(p-q)^{p+q}-1}-1\right)=\frac{(p+q)^{2q}-(p-q)^{2q}}{(p+q)^{p-q}(p-q)^{p+q}-1}\]is an integer.

The only pair that works is \((3,2)\). We can easily verify this is the only solution for \(q\le3\) by applying the following claim and finite-case checking:

Claim: \(p<2q\)

Proof. We know that \((p+q)^{p-q}(p-q)^{p+q}-1\) divides \((p+q)^{2q}-(p-q)^{2q}\). But if \(p\ge2q\), then \[(p+q)^{p-q}(p-q)^{p+q}-1\ge(p+q)^{2p-2q}-1\ge(p+q)^{2q}-(p-q)^{2q},\]absurd. \(\blacksquare\)

Henceforth assume \(q\ge5\). Then \(p+q\) and \(p-q\) are even, so define \[s:=(p+q)^{(p-q)/2}(p-q)^{(p+q)/2}-1.\]Let \(r\) be a prime dividing \(s^2-1\), so \(r\) divides \((p+q)^{p+q}(p-q)^{p-q}-1\).

Claim: Either \(r=q\) or \(q\) divides \(r-1\).

Proof. Recall that \(s^2-1\) divides \((p+q)^{2q}-(p-q)^{2q}\), so \[\left(\frac{p+q}{p-q}\right)^{2q}\equiv1\pmod r.\]It follows that \(\operatorname{ord}_r(\frac{p+q}{p-q})\in\{1,2,q,2q\}\).
  • If \(\operatorname{ord}_r(\frac{p+q}{p-q})\in\{q,2q\}\), then \(q\mid r-1\).
  • Otherwise \(\operatorname{ord}_r(\frac{p+q}{p-q})\in\{1,2\}\), and so \[r\mid\left(\frac{p+q}{p-q}\right)^2-1\mid(p+q)^2-(p-q)^2=4pq.\]
    • Since \(s^2-1\) is odd, \(r\ne2\).
    • If \(r=p\), then \[0\equiv s^2-1\equiv q^{2p}-1\equiv q^2-1\pmod p,\]so \(p\) divides either \(q-1\) or \(q+1\). This is absurd by \(p>q\).
    Hence \(r=q\).
\(\blacksquare\)

It follows that both \(s-1\) and \(s+1\) are either 0 or 1 modulo \(q\). This is impossible under our assumption that \(q\ge5\).
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IndoMathXdZ
696 posts
#10
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MarkBcc168 wrote:
Find all pairs $(p,q)$ of prime numbers which $p>q$ and
$$\frac{(p+q)^{p+q}(p-q)^{p-q}-1}{(p+q)^{p-q}(p-q)^{p+q}-1}$$is an integer.
Not so great of a problem but quite ok for N5.
\begin{align*}
(p + q)^{p - q}(p - q)^{p + q} - 1 &\mid (p + q)^{p + q} (p - q)^{p - q} - 1 \\
(p + q)^{p - q} (p - q)^{p + q} - 1 &\mid (p + q)^{p + q} (p - q)^{p - q} - (p + q)^{p - q} (p - q)^{p + q} \\
(p + q)^{p - q} (p - q)^{p + q} - 1 &\mid (p + q)^{p - q} (p - q)^{p - q} ( (p + q)^{2q} - (p - q)^{2q} ) \\
(p + q)^{p - q} (p - q)^{p + q} - 1 &\mid (p + q)^{2q} - (p - q)^{2q} 
\end{align*}Claim 01. $p < 3q$.
Proof. Notice that $(p + q)^{2q} > (p - q)^{2q}$, and by divisibility criteria, we have
\[ (p + q)^{p - q} - 1 \le (p + q)^{p - q}(p - q)^{p + q} - 1 \le (p + q)^{2q} - (p - q)^{2q} \le (p + q)^{2q} - 1 \]Therefore, we conclude that $p - q \le 2q$, but since $p$ and $q$ are both primes, we conclude that $p < 3q$.
Claim 02. If $q = 2$, then the only solution is $(3,2)$.
Proof. By the previous claim, we have $p \le 3q$, so we just need to check $(3,2)$ and $(5,2)$. Notice that $(3,2)$ obviously satisfies, and therefore $(3,2)$ is a solution.
Now, we may assume that $p  > q \ge 3$ are both odd, and therefore $p + q$ and $p - q$ are both even.
Claim 03. If $r \mid (p + q)^{p - q} (p - q)^{p + q} - 1$, then either $r = q$ or $r \equiv 1 \ (\text{mod} \ q)$.
Proof. We have
\[ r \mid (p + q)^{p - q} (p - q)^{p + q} - 1 \mid (p + q)^{2q} - (p - q)^{2q} \]Then, we have $o_r \left( \frac{p + q}{p - q} \right) \mid 2q$. Suppose that $q \nmid o_r \left( \frac{p + q}{p - q} \right)$, then we have \[ r \mid (p + q)^2 - (p - q)^2 = 4pq \]For obvious reason, $\gcd(r,4) = 1$ since $(p + q)^{p - q} - (p - q)^{p + q}$ is even. Now, we have $r = p$ or $r = q$.
If $r = p$, then
\[ p \mid (p + q)^{p - q} (p - q)^{p + q} -  1 \Rightarrow p \mid q^{2p} - 1 \]By Fermat Little Theorem, $p \mid (q^p)^2 - 1$, and therefore $p \mid q^2 - 1$. However, we know that $p \ge q + 2$ and $p \mid (q - 1)(q + 1)$. This is a contradiction.
We now have that for all primes $r$ such that
\[ r \mid (p + q)^{p - q} (p - q)^{p + q} - 1 \]then $r \equiv 0, 1 \ (\text{mod} \ q)$.
For me, this is the trickiest part of the solution, as everything else flows naturally: Let $N = (p + q)^{\frac{p - q}{2}} (p - q)^{\frac{p + q}{2}}$, which is an integer since $p - q$ and $p + q$ are both even. Therefore,
\[ r \mid N^2 - 1 \Rightarrow r \mid (N - 1)(N + 1) \]Now, we conclude that $\{ N - 1, N + 1 \} \equiv \{ 0, 1 \} \ (\text{mod} \ q)$. Now, notice that $(N + 1) - (N - 1) \in \{ 1, 0, -1 \} \ (\text{mod} \ q)$. This pretty much gives us $q \le 3$. Since we have $q \ge 3$, this forces $q = 3$.
Hence, from our first claim, we just need to check $p < 3q = 9$, which means we just need to check $(5,3)$ and $(7,3)$, both of which don't work.

Remark. This problem is really intimidating, yet the housekeeping (standard order stuffs to simplify things) did most of the tricks.
However, this thing is quite popular among order problems: that if you conclude that the prime divisors of $a$ is either $0$ or $1$ modulo $q$, then $a$ must be either $0$ or $1$ modulo $q$. Once you realized that $(p + q)^{p - q} (p - q)^{p + q} - 1 = N^2 - 1$ for some $N \in \mathbb{N}$, we are pretty much done.
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mathaddiction
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#11
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The answer is $(p,q)=(3,2)$, which obviously works.

We now show that it is the only solution. Notice that $p+q, p-q$ are both relatively prime with the denominator. As a result,
\begin{align*}
1&\equiv(p+q)^{p+q}(p-q)^{p-q}\\
&\equiv(p+q)^{p-q}(p-q)^{p+q}(p+q)^{2q}(p-q)^{2q}\\
&\equiv\left(\frac{p+q}{p-q}\right)^{2q}\pmod{(p+q)^{p-q}(p-q)^{p+q}-1}\\
\end{align*}This implies
$$(p+q)^{p-q}(p-q)^{p+q}-1|(p+q)^{2q}-(p-q)^{2q}\hspace{20pt}(1)$$In particular, we have $(p+q)^{p-q}\leq (p+q)^{2q}$, therefore, $p\leq 3q$, so if $q=2$ then the only solution is $p=3$. Now supppose $q$ is odd.

CLAIM 1. $p<2q$.
Proof.
Suppose $p\geq 2q$, then from $(1)$ we have
$$(p-q)^{p+q}\leq (p+q)^{3q-p}\leq (p+q)^q$$Let $p=kq$, then
$$(k-1)^{p+q}q^q\leq (k+1)^q$$Notice that $k<3$, therefore $q=3$, from which we easily check that there is no solution. $\blacksquare$

CASE I: $q|(p+q)^{p-q}(p-q)^{p+q}-1$.
Then $q|p^{2p}-1$. Notice that the $\text{ord}_q(p)|q-1$, therefore, it can only be $2$. This implies
$p\equiv \pm 1\pmod q$
Therefore, $p=2q-1$, the equation becomes
$$(3q-1)^{q-1}(q-1)^{3q-1}-1|(3q-1)^{2q}-(q-1)^{2q}$$hence $(q-1)^{3q-1}\leq (3q-1)^{q+1}$, but by easy induction we have $(q-1)^{3q-1}>(3q-1)^{q+1}$ for $q\geq 5$, contradiction.

CASE II: $q\nmid (p+q)^{p-q}(p-q)^{p+q}-1$
Suppose $r$ is a prime such that $r|(p+q)^{p-q}(p-q)^{p+q}-1$
CLAIM 2. If $r\neq p$ then $r\equiv 1\pmod q$
Proof.
Since $(p+q)^{p-q}(p-q)^{p+q}-1$ is odd, $r\neq 2$.
Notice that $r\neq p,q$ by assumption, moreover,
$$1\equiv\left(\frac{p+q}{p-q}\right)^{2q}\pmod r$$Let the order of $\displaystyle\left(\frac{p+q}{p-q}\right)$ modulo $r$ be $d$, notice that $r$ cannot be $2$ or otherwise $r|(p+q)^2-(p-q)^2$ which implies $r|4pq$, contradiction.
Therefore, $q|d$, hence $q|d|r-1$ as desired. $\blacksquare$

Notice that $p-q,p+q$ are both even, let $a=(p+q)^{\frac{p-q}{2}}(p-q)^{\frac{p+q}{2}}-1$, $b=(p+q)^{\frac{p-q}{2}}(p-q)^{\frac{p+q}{2}}+1$. By LTE,
$$v_p(ab)\leq v_p((p+q)^{2q}-(p-q)^{2q})=1$$Therefore, if $v_p(ab)=0$, then we have $a\equiv b\equiv 1\pmod q$, hence $q=2$ since $b-a=2$. Otherwise, one of $a,b$ is congruent to $1$ mod $q$ while the other is congruent to $p$ mod $q$, hence $p=q+3$ or $2q-1$, the former is impossible and the latter is rejected in CASE I.
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MatBoy-123
396 posts
#12 • 1 Y
Y by Mango247
Some orders and bounding is enough to solve...
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hydo2332
435 posts
#13
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Let $(p-q)^{p+q}(p+q)^{p-q} -1 = N$. Notice that the problem is equivalent to $(p+q)^{2q} - (p-q)^{2q} \equiv 0 \pmod N$ due to the fact that $(p,N) = (q,N)= 1$. And this implies that the order of the element $\frac{p+q}{p-q}$ is either $1,2,q,2q$ modulo $N (*)$. Now, we have $(p+q)^{p-q} < N \leq (p+q)^{2q} - (p-q)^{2q} < (p+q)^{2q}$, which means that $p < 3q$.
Claim. $p \leq 2q$ or $q =2$
Suppose that $2q < p < 3q$. We now have that $(3q)^{q}(q)^{3q} < (p+q)^{p-q}(p-q)^{p+q} - 1 \leq (p+q)^{2q} - (p-q)^{2q} < (4q)^{2q}
3^q \cdot q^{4q} < 4^{2q} \cdot q^{2q} \implies
3q^4 < 16q^2 \implies
3q^2 < 16 \implies
q = 2$ is the only possibility. Hence, the lemma follows.

Now, we divide into the cases of the orders, as said in $*$. Let $r$ be a prime dividing $N$. If the order is $1$ or $2$, we have that $r | 4pq$, and hence $r$ is either $2,p,q$. If $q=2$ we have that $p < 6$, so we just need to try the values. From now on, we assume $q \geq 3$, and hence $N$ is odd, ,so $r$ is either $p,q$. $r$ can't be $q$ because $p>q$ and can't be $p$ because $p \leq 2q$ by the claim.

Finally, if the order is $2q$ or $q$, we have that $2q | r-1$, which implies that every prime divisor of $N$ is of the form $1 + kq$ ,which implies $N \equiv 1 \pmod q$. But notice that $N$ is of the form $x^2 - 1$ for some $x$, and hence $(x-1)(x+1) \equiv 1 \pmod q$, which is not possible in this case.
Hence, the only solution is $(3,2)$, which comes from $q=2$ and trying $p < 3q$.
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tigerzhang
351 posts
#14 • 1 Y
Y by Bradygho
The answer is $(p,q)=(3,2)$, which works.

Let $r$ be an arbitrary prime divisor of $(p+q)^{p-q}(p-q)^{p+q}-1$. Then, we must have $(p+q)^{p+q}(p-q)^{p-q} \equiv 1 \pmod{r}$. Thus, $$\frac{(p+q)^{p+q}(p-q)^{p-q}}{(p+q)^{p-q}(p-q)^{p+q}}=\left(\frac{p+q}{p-q}\right)^{2q} \equiv 1 \pmod{r}.$$If $\operatorname{ord}_r\left(\frac{p+q}{p-q}\right)$ is $1$ or $2$, then $(p+q)^2 \equiv (p-q)^2$, so $4pq \equiv 0 \mod r$. Therefore, either $r$ is even, $r=p$, or $r=q$. Otherwise, $q \mid \operatorname{ord}_r\left(\frac{p+q}{p-q}\right)$, so $r \equiv 1 \pmod{q}$. We have three cases.

Case 1: $q=2$. Then, $(p+2)^{p-2}(p-2)^{p+2}-1 \mid (p+2)^{p+2}(p-2)^{p-2}-1$, so \[(p+2)^{p-2}(p-2)^{p+2}-1 \mid (p+2)^{p+2}(p-2)^{p-2}-(p+2)^{p-2}(p-2)^{p+2}.\]Thus, we have $$(p+2)^{p-2}(p-2)^{p+2}-1 \mid (p+2)^4-(p-2)^4.$$If $p \geq 7$, then we have a contradiction because of size. We obtain that $p=3$ gives the solution $(3,2)$ and $p=5$ fails due to size reasons. Now, assume henceforth that $p$ and $q$ are even.

Case 2: $(p+q)^{p-q}(p-q)^{p+q}-1$ is divisible by $p$. Then, we have $$q^{p-q}(-q)^{p+q}=(-1)^{p+q}q^{2p} \equiv 1 \pmod{p}.$$Since $p$ and $q$ are even, we have $q^{2p} \equiv 1 \pmod{p}$, so $\operatorname{ord}_p(q) \mid \gcd(p-1,2p)=2$. Thus, $p \mid q^2-1$, so $p \mid \frac{q-1}{2}$ or $p \mid \frac{q+1}{2}$, which fails because $p>q$.

Case 3: $(p+q)^{p-q}(p-q)^{p+q}-1$ is divisible by $p$. If $(p+q)^{p-q}(p-q)^{p+q}-1$ is divisible by $q$, then, we have $p^{2p}-1 \equiv 0 \pmod{q}$, so $\operatorname{ord}_p(q) \mid \gcd(q-1,2p)$. Since $p>q$, we see that $\gcd(q-1,2p)=2$. Thus, $q \mid p^2-1$, so $q \mid p-1$ or $q \mid p+1$. If $q \mid p-1$, the prime factorization of $(p+q)^{p-q}(p-q)^{p+q}-1$ must consist of $\pm 1 \mod{q}$ primes. Thus, we $(p+q)^{p-q}(p-q)^{p+q}-1 \equiv \pm 1 \pmod{q}$. Since $$(p+q)^{p-q}(p-q)^{p+q}-1=\left((p+q)^{\frac{p-q}{2}}(p-q)^{\frac{p+q}{2}}-1\right)\left((p+q)^{\frac{p-q}{2}}(p-q)^{\frac{p+q}{2}}+1\right),$$and both factors cannot be $1 \mod{q}$, we are done in this case. Otherwise $q \mid p+1$. We can do a similar Euclidean algorithm as in Case 1 to obtain that $p=2q-1$. Now, we are done by bounding.
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bora_olmez
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#15
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The difficulty is spotting the difference of squares, otherwise, I agree that this is rather straightforward for an N5.
Solution
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mijail
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#16
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My answer is $(p,q)=(3,2)$.
Suppose that $q>2$ so $p+q,p-q$ are eve, if we substract $1$ from the integer and remove the factors $p+q,p-q$ (we can do this because $p+q,p-q$ are coprimes with the denominator). We have that if $M = (p + q)^{p - q}(p - q)^{p + q} - 1$ then:
$$M \mid (p + q)^{2q} - (p - q)^{2q}$$The key idea is the following:
Key Claim: Let r be a prime divisor of $M$ then $r\equiv 1,0 \pmod{q}$
Proof: We have that: $$r \mid (p + q)^{2q} - (p - q)^{2q} \implies (\frac{p+q}{p-q})^{2q} \equiv 1 \pmod{r}$$Then $T= \operatorname{ord}_r\left(\frac{p+q}{p-q}\right)$ is divisor of $2q$ but also is divisor of $r-1$. But if $T$ is $q$ or $2q$ then $r\equiv 1 \pmod{q}$
And if $T$ is $1$ or $2$ then: $$(\frac{p+q}{p-q})^{2} \equiv 1 \pmod{r} \implies 4pq \equiv 0 \pmod{r}$$But $M$ is odd so $r=p$ or $r=q$.
If $r=p$ then: $$p \mid (p + q)^{p - q}(p - q)^{p + q} - 1 \implies p \mid (q)^{p - q}(- q)^{p + q} - 1  \implies p \mid q^{2p} - 1$$So by FLT: $p \mid q^{2} - 1   \implies p \mid (q-1)(q+1)$ then $p \leq q+1$ but $p>q$ and $p,q$ are odd so $p-q \ge 2$ (Contradiction!)

Then $r=q$ or $r\equiv 1 \pmod{q}$ $\square$.
Finally the claim implies that all divisor of $M$ is $\equiv 1,0 \pmod{q}$ but $M$ have divisors: $$M=((p + q)^{\frac{p - q}{2}}(p - q)^{\frac{p + q}{2}} - 1)((p + q)^{\frac{p - q}{2}}(p - q)^{\frac{p + q}{2}}+1)$$Then exists two divisors of $M$ with difference $2$ and both are $\equiv 1,0 \pmod{q}$ so the difference is $2 \equiv 1,0,-1 \pmod{q}$ then the unique possible value of $q$ is $3$.
Finally:
$$M \mid (p + q)^{2q} - (p - q)^{2q} \implies M\leq (p + q)^{2q} - (p - q)^{2q} \implies M \leq (p + q)^{2q} - 1 \implies (p + q)^{p - q}(p - q)^{p + q} \leq (p + q)^{2q}$$So we have: $(p - q)^{p + q} \leq (p + q)^{3q-p} \implies p<3q$. If $q=3 \implies p<9$ if $p=7  \implies 4^{10} \leq 100$ obvious this is false.
If $q=3,p=5$ then n the problem statement we replace: $ 2^{14}-1 \mid 2^{26}-1 \implies 14 \mid 26$ that is false.
If $q=2$ then $p<6$ if $p=5  \implies 3^{7}\leq 7$ is false.
Then the unique possible pair is $(p,q)=(3,2)$ that works.
This post has been edited 1 time. Last edited by mijail, Jan 24, 2022, 6:01 PM
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NoctNight
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#17 • 1 Y
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Solution
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HamstPan38825
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#18
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The only pair is $(p, q) = (3, 2)$.

Assume first that $p, q$ are both odd primes. Then, let some prime $r \mid (p+q)^{p-q}(p-q)^{p+q} - 1$. Then, using the divisibility condition and dividing, we have $$\left(\frac{p+q}{p-q}\right)^{2q} \equiv 1 \pmod r.$$Thus, either $q \mid r-1$ or $$(p+q)^2 \equiv (p-q)^2 \pmod r \iff r \mid 4pq.$$If $r = p$, then $$1 \equiv (p+q)^{p-q}(p-q)^{p+q} \equiv q^{2p} \pmod p,$$hence $p \mid q^{2p} - 1$, or $p \mid q^2 - 1$. But this is obviously impossible as $p, q$ are both odd primes.

As a result, either $r=2$, which is impossible, or $r = q$. This crucially means that $(p+q)^{p-q}(p-q)^{p+q} - 1$ is the product of $q$ and $1 \bmod q$ primes. On the other hand, $$(p+q)^{p-q}(p-q)^{p+q} - 1 = ((p+q)^{(p-q)/2}(p-q)^{(p+q)/2}+1)((p+q)^{(p-q)/2}(p-q)^{(p+q)/2}-1),$$so each of these factors is $0$ or $1$ mod $q$. But this clearly implies $q \leq 3$.

Assuming that $q = 3$, we can reduce to a finite case check by showing that $p < 2q$. However, this is evident by size reasons, as otherwise $$\gcd((p+q)^{p-q}(p-q)^{p+q} - 1, (p+q)^{p+q}(p-q)^{p-q} - 1) = \gcd((p+q)^{p-q}(p-q)^{p+q} - 1, (p+q)^{2q} - (p-q)^{2q}).$$When $p>2q$, the former expression is always greater, but this contradicts divisibility.

Now, if $q=2$, testing cases and using the previous conclusion, we obtain $(3, 2)$ is the only solution.
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awesomeming327.
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#19
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Let $a=p+q$ and $b=p-q$. We have
\begin{align*} 
a^bb^a-1 &\mid a^ab^b-1 \\
a^bb^a-1 &\mid a^ab^b-a^bb^a \\
a^bb^a-1 &\mid a^bb^b(a^{a-b}-b^{a-b}) \\
a^bb^a-1 &\mid a^{2q} - b^{2q}
\end{align*}Thus, $\left(ab^{-1}\right)^{2q} \equiv 1\pmod {a^bb^a-1}$. Let $r$ be the smallest prime dividing $a^bb^a-1$ then $g=\text{ord}_r(ab^{-1})\mid 2q$ and $g\mid r-1$.

Suppose $q\ge 5$. Clearly, $r\neq 2$. If $g=1$ then $a\equiv b\pmod r$. If $g=2$ then $a\equiv -b\pmod r$. In the first case, $r\mid 2q$, so $r=q$. In the second, $r=p$. $r=p$ is clearly impossible since \[a^bb^a-1\equiv q^{p-q}q^{p+q}-1\equiv q^{2p}-1\equiv q^2-1\pmod p\]and $(q-1)$, $(q+1)$ are both coprime to $p$ since $p$ is a larger prime than $q$. If $g=q$ or $2q$ then $q\mid r-1$. Thus, the only prime factors of $a^bb^a-1$ are $0$ or $1 \pmod q$. Thus, any factor of $a^bb^a-1$ is also $0$ or $1\pmod q$. This is absurd, since $a^bb^a-1$ is one less than a perfect square and thus factors as $(t-1)(t+1)$ for some $t$, and one of them must break the $0$ or $1\pmod q$ rule.

Suppose $q=3$. We have $a$ and $b$ are even, and so \[a^bb^a-1\mid a^6-b^6\implies a^bb^a-1\mid (a/2)^6-(b/2)^6\]If $p>9$ then $2b>a$ so surely $(a/2)^6-(b/2)^6 < (a/2)^6 < b^6 < b^a <a^bb^a-1$ which implies $a=b$, absurd. Thus, $p=7$ or $p=5$. It can be verified that neither works.

Suppose $q=2$. We have $a$ and $b$ are both odd. We also have \[a^bb^a-1\mid a^4-b^4\]If $p>6$ then $b>4$ so $a^4-b^4<a^4\le a^b<a^bb^a-1$ and so $a=b$, again absurd. Thus, $p$ is $3$ or $5$, and it can easily be verified that $(3,2)$ is the only solution.
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Leo.Euler
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#20
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For notational purposes, let $a = p+q$ and $b=p-q$. Observe that $a^bb^a \equiv 1 \pmod{a^bb^a - 1}$, from which the divisibility is equivalent to \[ (a^bb^a-1) \mid a^{2q}-b^{2q}. \]Let $p_0$ be an odd prime divisor of $a^bb^a-1$. Then \[ \left(\frac{a}{b}\right)^{2q} \equiv 1 \pmod{p_0}, \]so the order of $a/b$ modulo $p_0$ is one of $1$, $2$, $q$, and $2q$. We proceed to do casework on these possible orders.
  • If the order is divisible by $q$, then $p_0 \equiv 1 \pmod{q}$.
  • If the order is $1$ or $2$, then in either case we have \[ (p+q)^2 \equiv (p-q)^2 \pmod{p_0}, \]from which $p_0 = p$ or $p_0 = q$.
If $p_0=p$, then \[ q^{2p} \equiv 1 \pmod{p}, \]so $p \mid q-1$ or $p \mid q+1$ by FLT. Neither are possible since $p>q$. Thus, each prime divisor of $a^bb^a-1$ is one of $q$ and $1 \pmod{q}$. In particular, $(a^bb^a-1)/q^{\nu_q(a^bb^a-1)}$ is $1 \pmod{q}$.

Notice that we can write \[ p_0 \mid (a^{b/2}b^{a/2}-1)(a^{b/2}b^{a/2}+1) \]and realize that $p_0$ must divide exactly one of the two factors. Since each factor is in $\{0, 1\} \pmod{q}$, and their difference is $2$, we must have $q=2$ or $q=3$. To check which $p$ work along with these values of $q$, we first bound $p$.

Claim: We have $p \le 2q$.
Proof. Recall that $(a^bb^a-1) \mid a^{2q}-b^{2q}$ from our initial discussion. Then show that the LHS is greater than the RHS when $p > 2q$, so $p \le 2q$.
:yoda:

Checking the resulting cases by hand returns $(3, 2)$ as the only solution.
This post has been edited 1 time. Last edited by Leo.Euler, Nov 30, 2023, 11:50 PM
Reason: typo
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math_comb01
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#21
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anantmudgal09 wrote:
Quotes on this at PDC:
"Remember kids, $n^2-1$ factors as $(n-1)(n+1)$."
"Primes are odd". "Uh, except $2$ right?" "No one cares [insert person name]".
Posting for comedic reasons,
The key is to notice that $p^2-q^2=(p-q)(p+q)$
Notice then the equation rewrites as $(p+q)^{p-q}(p-q)^{p+q}-1 \mid (p+q)^{2q}-(p-q)^{2q}$ this also gives us the $p \leq 3q$ bound/ Let $a=p+q,b=p-q$
Claim:(Characterization of primes dividing $a^bb^a-1$) Every prime divisor $r$ of $a^bb^a-1$ must be of form $qk+1$ unless $p \equiv \pm 1 (\mod q)$
Proof
Suppose for now that every prime divisor is of form $qk+1$ then get a contradiction.
then use some inequalities for $p = 2q \pm 1$ to get the only solution $(3,2)$
This post has been edited 1 time. Last edited by math_comb01, Jan 25, 2024, 7:39 PM
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thdnder
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Answer: $(p, q) = (2, 3)$ is the only solution.

Firstly, we'll prove that there is no solution if $q > 3$. Assume $q > 3$. Let $T = (p+q)^{p-q}(p-q)^{p+q} - 1$. Then note that $(p+q)^{p+q}(p-q)^{p-q} \equiv 1 (T)$ and $(p+q)^{p-q}(p-q)^{p+q} \equiv 1 (T)$, so $\frac{(p+q)^{p+q}(p-q)^{p-q}}{(p+q)^{p-q}(p-q)^{p+q}} = (\frac{p+q}{p-q})^{2q} \equiv 1 (T)$.

Since $T = (p+q)^{p-q}(p-q)^{p+q} - 1 = ((p+q)^{\frac{p-q}{2}}(p-q)^{\frac{p+q}{2}} - 1)((p+q)^{\frac{p-q}{2}}(p-q)^{\frac{p+q}{2}} + 1)$ and $(p+q)^{\frac{p-q}{2}}(p-q)^{\frac{p+q}{2}} - 1 \equiv p^p - 1 (q)$, $(p+q)^{\frac{p-q}{2}}(p-q)^{\frac{p+q}{2}} + 1 \equiv p^p + 1 (q)$, hence either $p^p + 1 \not\equiv 0, 1 (q)$ or $p^p - 1 \not\equiv 0, 1 (q)$. Therefore, there exists a prime $r$ such that $r \mid T$ and $r \not\equiv 0, 1 (q)$. But we have $(\frac{p+q}{p-q})^{2q} \equiv 1 (T)$, which means $((\frac{p+q}{p-q})^{2q} \equiv 1 (r))$, so $(\frac{p+q}{p-q})^r \equiv 1 (r)$, or equivalently, $r \mid 4pq$, a contradiction.

Now assume $q \le 3$. We split the problem into 2 cases:

Case 1: $q = 2$.

In this case, we have $(p+2)^{p-2}(p-2)^{p+2} - 1 \mid (p+2)^4 - (p-2)^4 = 16p(p^2 + 4)$. If $p \ge 5$, then $(p+2)^{p-2}(p-2)^{p+2} - 1 \ge (p+2)^3(p-2)^{p+2} - 1 \ge (p+2)^3 \cdot 2187 - 1 > 16p(p^2 + 4)$, so $p \le 3$, which forces $p = 3$ and we can easily check that $(2, 3)$ is a solution.

Case 2: $q = 3$.

If $p \ge 11$, then $(p+3)^{p-3}(p-3)^{p+3} - 1 \ge (p+3)^8 \cdot 5^11 - 1 > 36p(p^4 + 30p^2 + 81) = (p+3)^6 - (p-3)^6$, so $(p+3)^{p-3}(p-3)^{p+3} - 1 \nmid (p+3)^6 - (p-3)^6$, a contradiction. Hence $p$ equals $5$ or $7$ and one can check that $(3, 5)$ and $(3, 7)$ are not solutions. Hence we're done. $\blacksquare$
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shendrew7
818 posts
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Our only solution is $\boxed{(3,2)}$. Note that $q=2,3$ is just a finite case check due to size issues, giving us our claimed solution. We now assume $p,q \ge 5$, and suppose $r$ is an arbitrary prime divisor of the denominator, which also must divide the numerator. We then require
\[(p+q)^{p+q} (p-q)^{p-q} \equiv (p+q)^{p-q} (p-q)^{p+q} \equiv 1 \pmod r\]\[\implies \left(\frac{p+q}{p-q}\right)^{2q} \equiv 1 \pmod r \implies k := \operatorname{ord}_r \left(\frac{p+q}{p-q}\right) \mid \gcd(2q,r-1) \mid 2q.\]
First notice $r=2$ and $r=p$ both result in contradictions. Then
\begin{align*}
k=1,2 &\implies (p+q)^2 \equiv (p-q)^2 \pmod r \implies r=q \\
k=q,2q &\implies \gcd(2q,r-1) = q,2q \implies r \equiv 1 \pmod q.
\end{align*}
We finish by using difference of squares on the numerator to get two factors differing by 2 which must both be $0,1 \pmod r$, contradiction. $\blacksquare$
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megarnie
5718 posts
#24
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The only solution is $(3,2)$, which works. Now we prove everything else fails. Suppose $(p,q)$ isn't $(3,2)$.

Let $a = p+q, b = p-q$. Clearly $b > 1$.

We have $a^b b^a - 1 \mid a^a b^b - 1$. Hence\[a^b b^a - 1 \mid a^a b^b - 1 - (a^b b^a - 1) = a^a b^b - a^b b^a\]Since $a,b$ are relatively prime to $a^b b^a$, we can divide the RHS by $(ab)^b$, so\[ a^b b^a - 1 \mid a^{a-b} - b^{a-b} = a^{2q} - b^{2q}\]Since $a > b$, we have that $a^b b^a - 1 < a^{a-b} - b^{a-b} < a^{a-b}$. Now since $b > 1$, $a^b b^a - 1 > a^b$, so $a^b < a^{a-b}$, meaning $a > 2b$. Hence $p + q > 2p - 2q$, so $p < 3q$.

Case 1: $q >3$. Then $p > q \ge 5$.

Claim: $a^b b^a - 1$ has some prime factor that is not $1\pmod q$.
Proof: Suppose otherwise. Then all divisors of $a^b b^a - 1$ are $1\pmod q$. Then notice that $a,b$ are both even, so $a^b b^a$ is a perfect square. Let $x^2 = a^b b^a$, we have that $(x-1)(x+1) = a^b b^a - 1$, so $x -1$ and $x + 1$ are $1\pmod q$, implying $q = 2$, contradiction. $\square$.

Let $r$ be an odd prime dividing $a^b b^a - 1$ that is not $1\pmod q$. We have $r$ divides $a^{2q} - b^{2q}$, so $\left( \frac ab \right)^{2q} \equiv 1\pmod r$.

Hence the order of $\frac ab$ modulo $r$ divides $2q$. However, it can't be a multiple of $q$ as $q$ doesn't divide $r - 1$. Hence the order of $\frac ab$ modulo $r$ is either $1$ or $2$.

If the order of $\frac ab$ modulo $r$ is $2$, then $r\mid a^2 - b^2 = (a-b)(a+b)$. But $\frac ab \not \equiv 1\pmod r$, so $r$ divides $a + b = 2p$, meaning that $r = p$. We have $a \equiv q\pmod p$ and $b\equiv -q\pmod p$, and both $a,b$ are even, so $a^b b^a \equiv q^b \cdot q^a \equiv q^{2p} \pmod p$, so $q^{2p} \equiv 1\pmod p$. But $q^p \equiv q\pmod p$, so $q^{2p} \equiv q^2 \pmod p$, meaning that $q \in \{-1,1\}\pmod p$. Since $q < p$, this means $q = p - 1$, which is absurd since $ q > 2$. Therefore, the order of $\frac ab$ modulo $r$ must be $1$.

Then $r \mid a - b$, so $r = q$. Hence $a^b  b^a \equiv 1\pmod q$. However, we have $a \equiv b \equiv p \pmod q$, so $p^{a+b} = p^{2p} \equiv 1 \pmod q$. If the order of $p$ modulo $q$ is $p$ or $2p$, we have a contradiction as $p$ cannot divide $q - 1$ (since $p > q$). Since the order of $p$ mod $q$ divides $2p$, it must be $1$ or $2$. Hence $p^2 \equiv 1\pmod q$, so $p\equiv \pm 1 \pmod q$.

Since $q < p < 3q$, $p\in \{q+1, 2q-1, 2q+1, 3q-1\}$. By parity, $p$ can't be $q+1$ or $3q - 1$.

If $p = 2q - 1$, then $a = 3q - 1$, $b = q - 1$, so $(3q - 1)^{q - 1} (q-1)^{3q - 1 } - 1$ divides $(3q - 1)^{2q} - (q-1)^{2q}$.

Hence $(3q-1)^{q-1} \cdot (q-1)^{3q - 1} $ must be less than $(3q-1)^{2q}$, implying that $(q-1)^{3q - 1} < (3q - 1)^{q+1}$.

However, we have\begin{align*}
(3q - 1)^{q+1} < (3q - 3)^{q+1} = 3^{q+1} (q-1)^{q+1}   \\
< (q-1)^{q+1} (q-1)^{q+1} = (q-1)^{2q+2} \\ 
< (q-1)^{3q - 1}, \\
\end{align*}absurd.

Therefore, $p = 2q + 1$. But then $a = 2q + 1, b = q  + 1$, so\[(3q + 1)^{q+1} (q+1)^{3q + 1} - 1 \mid (3q + 1)^{2q} - (q+1)^{2q}\]Hence $(q+1)^{3q + 1} (3q + 1)^{q+1}  < (3q+1)^{2q}$, so $(q+1)^{3q + 1} < (3q + 1)^{q-1}$.

But then\begin{align*}
(3q+1)^{q-1} < (3q+3)^{q-1}  = 3^{q-1} (q+1)^{q-1} \\
< (q + 1)^{2q - 2} < (q+1)^{3q  + 1}, \\
\end{align*}absurd.

Case 2: $q \le 3$.
If $q = 2$, then $p < 3 \cdot 2 = 6$, so $p = 3$ or $ p =5$. To see $p = 5$ does not work, notice it implies $a = 7$ and $b = 3$, and $7^3 \cdot 3^7 - 1$ clearly does not divide $7^4 - 3^4$ (in fact it is much larger). Hence $p = 3$.

If $q = 3$, then $5 < p < 3\cdot 3 = 9$, so $p \in \{5,7\}$. If $p = 5$, then $a = 8, b = 2$, but $8^2 \cdot 2^8 - 1$ doesn't divide $8^6 - 2^6$ (this can be proven by dividing the RHS by $2^6$ and getting a size contradiction). If $p = 7$, then $a  = 10, b =4$, but $10^4 \cdot 4^{10} $ doesn't divide $10^6 - 4^6$ (since $10^4 \cdot 4^10 > 10^6$). Hence $q = 3$ is not possible.





Therefore, we must have $p = 3, q = 2$, so we are done.
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AR17296174
39 posts
#25
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Nice problem!
Though really straightforward for N5. Here is my solution!
Attachments:
2017N5.pdf (149kb)
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Ilikeminecraft
734 posts
#26
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It suffices to evaluate when $\frac{(p + q)^{2q} - (p - q)^{2q}}{(p + q)^{p - q}(p - q)^{p + q} - 1}$ is integer. Suppose $r$ is a prime such that $r\mid (p + q)^{p - q}(p - q)^{p + q} - 1.$ Hence, $r\mid (p + q)^{2q}-(p - q)^{2q}.$ We obviously require $2q > p - q$ due to size, or $p < 3q.$ We also have the equivalent condition \[\left(\frac{p + q}{p - q}\right)^{2q} \equiv 1\pmod r.\]
We have 3 cases:
\begin{enumerate}
\item $\operatorname{ord}_r\left(\frac{p + q}{p - q}\right) = 1,$ then $\frac{p + q}{p - q} \equiv 1\pmod r,$ so then $2q \equiv 0\pmod r,$ so $r = q.$
\item $\operatorname{ord}_r\left(\frac{p + q}{p - q}\right) = 2,$ then $\frac{p + q}{p - q} \equiv 1\pmod r,$ so then $2p \equiv 0\pmod r,$ so $r = p.$ This implies that $q^2-1\equiv0\pmod p.$ Then, $p = q + 1.$ It isn't hard to verify that $p = 3, q = 2$ is a valid solution.
\item If $q\mid\operatorname{ord}_r\left(\frac{p + q}{p - q}\right),$ then $r\equiv1\pmod q$
\end{enumerate}
Thus, all prime factors of $(p + q)^{p - q}(p - q)^{p + q} - 1$ are exactly $q$ or $1\pmod q.$ However, $((p + q)^{\frac{p - q}{2}}(p - q)^{\frac{p + q}{2}} - 1)((p + q)^{\frac{p - q}{2}}(p - q)^{\frac{p + q}{2}} + 1)$ have distinct values modulo $q$ unless $q = 3.$ It is easy to verify $p = 5, 7$ by hand.
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