Israel Number Theory

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#1 h Jul 13, 2022, 2:55 AM • 17 Y

Y by dangerousliri, rightways, ylt_chn, GoodMorning, S.Ragnork1729, Mathlover_1, Matherer9654, mathmax12, NO_SQUARES, deplasmanyollari, buddyram, un_educated, Funcshun840, cubres, MS_asdfgzxcvb, PikaPika999, farhad.fritl

Find all triples $(a,b,p)$ of positive integers with $p$ prime and $a^p=b!+p.$

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squareman

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squareman

#2 h Jul 13, 2022, 2:56 AM • 30 Y

Y by fuzimiao2013, asdf334, YaoAOPS, khina, teasaffrontaffy, Bradygho, ike.chen, rightways, AndreiVila, megarnie, JG666, ylt_chn, rama1728, haha-SuperDaivish, ApraTrip, Quidditch, rayfish, A_Crabby_Crab, Mathlover_1, mathmax12, myh2910, Lightosv, Sedro, cubres, Funcshun840, Alex-131, MS_asdfgzxcvb, bakkune, PikaPika999, bin_sherlo

The answers are $\boxed{(2,2,2)}$ and $\boxed{(3,4,3)}.$

First verify $1 < (1+p)^{1/p} < 2$ for any prime $p$ so $b = 1$ impossible.

If $2\le b < p$ then take any prime $q\mid b!+p = a^p.$ Then $q > b.$ But $b! + p \le b^p < q^p \le a^p$ when $2 \le b < p,$ impossible.

If $b \ge p$ then $a = kp$ for some positive integer $k.$ Note $v_p(a^p - p) = 1$ so $b < 2p.$ Note $k < p$ to allow $(kp)^p \le (2p-1)! + p.$ If some prime $q < p$ divides $k$ then $b < q,$ but $(kp)^p > (q-1)! + p,$ contradiction. So $k = 1.$

Note that $v_2(p^p-p) \le 2\log_2(p-1)$ (LTE) and $v_2(p!) \ge p-\log_2(p+1)$ (Legendre's). So $p-\log_2(p+1) \le 2\log_2(p-1) \implies p < 10.$ Finite case check on $p=2,3,5,7$ finishes. $\blacksquare$

This post has been edited 2 times. Last edited by squareman, Jul 13, 2022, 3:02 AM

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Leartia

93 posts

Leartia

#3 h Jul 13, 2022, 3:06 AM • 4 Y

Y by angelthirty, Mango247, cubres, PikaPika999

Motivation: I have seen 2018 P5, 2019 P4. Think about size in addition to $v_p.$

Suppose $p>10.$
$\textit{(i)}$
If $b\geq 2p$ then $v_p(b!)\geq 2$ therefore $v_p(b!+p)=1$ hence $v_p(a^p)=1.$
This means that $p\mid a^p$ which implies that $p\mid a$ therefore $v_p(a^p)=pv_p(a)=p.$
Hence $p=1$ but this is absurd.
$\textit{(ii)}$
If $b<p.$ In this case $b!<b^p\Rightarrow b!+b<b^p+p\Rightarrow a^p<b^p+p.$
Since perfect $p$ powers should be at least $p+1$ apart we have that $a^p<b^p\Rightarrow a<b.$
This implies that $a\mid b!$ therefore $a\mid p$ and either $a=1$ or $a=p.$
$a=1$ doesn't work. If $a=p$ then $p^p=b!+p$ so a size argument gives an immediate contradiction.

$\textit{(iii)}$ If $p\leq b<2p.$
If there exists a prime $q$ such that $q\neq p$ and $q\mid a$ then $q>b\geq p.$
Furthermore, $a^p\geq q^pp^p>p^{2p}>(2p-1)!+p>b!+p.$
Hence no such $q$ can exist. Therefore $a=p^\alpha.$ By checking this size argument you see that we must have $\alpha=1.$ Thus $a=p.$
Using LTE we have $v_2(p^{p-1}-1)=v_2(p-1)+v_2(p+1)+v_2(p-1)-1=v_2(b!).$
Using that legendre formula or whatever it's called we get $v_2(b!)=b-s_2(b).$
Since we have $p\leq b<2p$ we get $v_2(b!)=b-s_2(b)\geq p-\log_2(2p).$
On the other hand $2\log_2(p-1)+\log(p+1)-1\geq 2v_2(p-1)+v_2(p+1)-1.$
Putting this stuff together we get $2\log_2(p-1)+\log_2(p+1)-1\geq p-\log_2(2p)$ But this does not hold for $p>10.$ Thus we get our desired contradiction.
Now we check $p=2,3,5,7$ with $b$ in the appropriate range.

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MarkBcc168

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MarkBcc168

#4 h Jul 13, 2022, 3:08 AM • 5 Y

Y by yaron235, GeoKing, 1475393141xj, cubres, AnthonyDraude

Very easy for a 5. I found this comparable to P4.

All answers are $(a,b,p) = (2,2,2), (3,4,3)$ . The key claim is the following.

Claim: $a$ is not divisible by any prime $q<p$ .

Proof: Assume so. Then, $q\nmid a^p-p=b!\implies b<q$ . Therefore,
$b! \leq (q-1)! < q^{p-1} < q^p-p,$ so $a < q$ , a contradiction. $\blacksquare$

Now we have two cases:

If $p \nmid a$ , then $p\nmid b!\implies b < p$ , so
$b! \leq (p-1)! < p^{p-1} \leq p^p-p,$ or $a<p$ . This gives a contradiction to the above claim.
If $p\mid a$ , then $p^2\nmid a^p-p \implies b < 2p$ , so
$\begin{align*} b! &\leq (2p-1)! \\ &= (1(2p-1))(2(2p-2)) \dots ((p-1)(p+1)) p \\ &< (p^2)(p^2)\dots (p^2) p \\ &= p^{2p-1} \leq (p^2)^p-p, \\ \end{align*}$ so $a < p^2$ . Therefore, the claim forces $a=p$ . If $p\geq 5$ , then by Zsigmondy theorem, we can find a prime $q$ such that $\operatorname{ord}_q(p) = p-1$ , so that $q\geq 2(p-1)+1 = 2p-1$ and $q\mid p^p-p$ . This forces $p^p-p=(2p-1)!$ , and we can finish by bounding because the LHS is too small.

In conclusion, we must have $a=p$ and $p\in\{2,3\}$ .

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asdf334

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asdf334

#5 h Jul 13, 2022, 3:11 AM • 2 Y

Y by ItalianPhysicist, cubres

squareman wrote:

The answers are $\boxed{(2,2,2)}$ and $\boxed{(3,4,3)}.$

First verify $1 < (1+p)^{1/p} < 2$ for any prime $p$ so $b = 1$ impossible.

If $2\le b < p$ then take any prime $q\mid b!+p = a^p.$ Then $q > b.$ But $b! + p \le b^p < q^p \le a^p$ when $2 \le b < p,$ impossible.

If $b \ge p$ then $a = kp$ for some positive integer $k.$ Note $v_p(a^p - p) = 1$ so $b < 2p.$ Note $k < p$ to allow $(kp)^p \le (2p-1)! + p.$ If some prime $q < p$ divides $k$ then $b < q,$ but $(kp)^p > (q-1)! + p,$ contradiction. So $k = 1.$

Note that $v_2(p^p-p) \le 2\log_2(p-1)$ (LTE) and $v_2(p!) \ge p-\log_2(p+1)$ (Legendre's). So $p-\log_2(p+1) \le 2\log_2(p-1) \implies p < 10.$ Finite case check on $p=2,3,5,7$ finishes. $\blacksquare$

:D

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Skender

30 posts

Skender

#7 h Jul 13, 2022, 4:58 AM • 2 Y

Y by Mango247, cubres

The solutions are $(2,2,2)$ are $(3,4,3)$ .

1. $p|a$ :
If not, then $b<p$ , by $\mod p$ , and $b<a$ , by $\mod a$ . Then, $a^p\geq a^{b+1}>2b!$ and $a^p\geq 2^p\geq 2p$ , so $a^p$ is too large.

2. $p\leq b\leq 2p-1$ :
$b!$ must be a multiple of $p$ , but not of $p^2$ , which is the same as $p\leq b\leq 2p-1$ .

3. $a=p$ :
Let $a=pk$ . If $1<k<p$ , then, by $\mod k$ , $k|p$ , contradiction. If $k\geq p$ , then $(p^2)^p\leq a^p=b!+p\leq (2p-1)!+p\leq (2p-1)!\cdot p=\prod_{i=0}^{p-1} (p-i)(p+i)< (p^2)^p$ , contradiction. Thus, $k=1$ .

4. $p<5$ :
Suppose otherwise. Then, $p^p>p!+p$ , so $b\geq p+1$ . Now, our equation is $p(p^{p-1}-1)=b!$ . But, by Lifting the Exponent Lemma, $v_2(p(p^{p-1}-1))=v_2(p-1)+v_2(p+1)+v_2(p-1)-1$ . On the other hand, letting $p-1=d_1d_2$ (since $p\geq 5$ implies $p-1$ not prime), we have $v_2(b!)\geq v_2(d_1)+v_2(d_2)+v_2(p-1)+v_2(p+1)= v_2(p-1)+v_2(p-1)+v_2(p+1)> v_2(p(p^{p-1}-1))$ , contradiction.

Thus, $a=p=2$ or $a=p=3$ , giving us the mentioned solutions.

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juckter

324 posts

juckter

#8 h Jul 13, 2022, 5:11 AM • 4 Y

Y by khina, wangqing_cn, DrYouKnowWho, cubres

This may actually be the single least interesting-looking IMO problem I've read so that's fun.

$(2, 2, 2)$ and $(3, 4, 3)$ . Discard $a = 1$ . Notice that $\gcd(a, b!) \mid p$ . Consider two cases depending on whether $p$ divides $a$ .

If $p \nmid a$ then $p \nmid b! + p$ so $b < p$ . Now if $a \le b$ then $a$ and $b!$ share some prime factor which must be less than $p$ , impossible. Therefore $a \ge b + 1$ and $a^p \ge (b + 1)^p > b^p + pb^{p - 1} > b! + p$ so we fail in this case.

If $p \mid a$ then $p \mid b! + p$ so $b \ge p$ . Moreover $p^2 \nmid a^p - p$ so $b < 2p$ . Now let $a = pk$ . Since $a$ cannot have prime factors in common with $b!$ other than $p$ we must either have $k = 1$ or $k \ge p$ . In the latter case we have $a^p - p \ge p^{2p} - p > (2p - 1)! \ge b!$ which is a contradiction, so we must have $k = 1$ and $a = p$ . We can also check $p^{2p} - p > p!$ so $b \ge p + 1$ .

Seeking a final contradiction we look at $\nu_2$ . By LTE we have $\nu_2(p^p - p) = 2\nu_2(p - 1) + \nu_2(p + 1) - 1$ . If $p \ge 7$ then from $b \ge p + 1$ we get

$\nu_2(b!) \ge \nu_2((p + 1)!) \ge \nu_2(p + 1) + \nu_2(p - 1) + \nu_2((p - 1)/2) + \nu_2(2) = 2\nu_2(p - 1) + \nu_2(p + 1)$
which contradicts $b! \mid a^p - p$ . We are then left to check $p = 2, p = 3$ and $p = 5$ which give the aforementioned solutions.

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VicKmath7

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VicKmath7

#10 h Jul 13, 2022, 6:07 AM • 1 Y

Y by cubres

Really doable NT, and the ideas are very similar to those in 2021 N5 as well. Here is my solution:
Case 1. Let $1b$ and $b!+p \leq b^p < a^p$ , done (the first inequality is due to $b<p$ ).
Case 2. Let $p \leq b <2p$ (the other cases are impossible, since the RHS will have $v_p=1$ ). Obviously $p|a$ , so let $a=cp$ . The idea is that $c=1$ is the only possibility.
Firstly, we can bound $c<p$ (we need this in order to get a prime $q \leq b$ divdiding $a$ , so as to finish as in the above case), otherwise $(2p-1)!+p=a^p \geq p^{2p}$ , but this is impossible, since $(2p-1)! \leq (\frac {2p-1}{2})^{2p-1} <p^{2p-1}$ by AM-GM, absurd (by the way, a similar bound with AM-GM was used in 2021 N5 to bound a factorial). Now take a prime $q|c$ and note that since $q<p<b$ , $q$ will divide $b!$ , hence $p$ , absurd. So the only left possibility is $c=1$ .
We have the equation $p(p^{p-1}-1)=b!$ . The idea is to use $v_2$ and Legendre (well, $v_2$ is easier to bound than $v_q$ for a prime $q|p-1$ ). Note that $v_2(b!)>\frac{b}{2}-1>\frac{p-1}{2}$ and $v_2(p^{p-1}-1)=2v_2(p-1)+v_2(p+1)-1<2log_2(p-1)$ , and now finite case check shall suffice.

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Assassino9931

1390 posts

Assassino9931

#12 h Jul 13, 2022, 6:30 AM • 2 Y

Y by Dhruv777, cubres

It was about time that an application of the Lifting the Exponent Lemma appears at IMO! (And indeed, as VicKmath7
pointed out, this feels too much at home if ISL 2021 N5 is fresh in your mind.)

If $p=2$ , then $b\geq 3$ is impossible by mod $3$ and only $a=b=2$ works. If $p=3$ , then $b\geq 7$ is impossible by mod $7$ and only $a=3$ , $b=4$ works. From now on, let $p\geq 5$ . We shall repeatedly use the bound $n! \leq (\frac{n}{2})^n$ for $n\geq 6$ which follows by induction since $n+1 \leq \frac{(n+1)^{n+1}}{2n^n} \Leftrightarrow (1+\frac{1}{n})^n \geq 2$ holds by opening the brackets on the left and taking only the first two terms into account.

Now suppose $p$ does not divide $a$ . Clearly $a=1$ is impossible. Observe that $p$ does not divide $b!$ and so $b<p$ , whence $a^p \leq (p-1)! + p < p!$ and so $a<p$ . If $b=1$ , then $1 = a^p - p \geq 2^p - p \geq 2$ , contradiction, so $b\geq 2$ . Hence if $a < b$ , then $a$ would divide both $a^p$ and $b!$ and hence $p$ , contradiction as $p$ is prime. So $a>b$ , thus (since $p>b \geq 2$ ) $b! > b^p - p > b^b - b$ (as $b^b(b^{p-b}-1) > 2^{p-b} - 1 \geq p-b$ ) and $(b-1)! > b^{b-1} - 1$ which is false since $b^{b-1} \geq (b-1)!$ for $b\geq 2$ (both sides have the same number of multipliers, the left-hand one's all are greater). Hence there are no solutions when $p$ does not divide $a$ .

Next, consider $a = p^kx$ where $p$ does not divide $x$ and $k\geq 1$ . Then $p^{kp}x^p = b! + p$ . If $b\geq 2p$ , then $b!$ and the left-hand side would be divisible by $p^2$ but $p$ would not be, so impossible. Hence $b<2p$ and so (using the above factorial bound) $b! + p \leq (2p-1)! + p \leq (\frac{2p-1}{2})^{2p-1} + p < p^{2p-1} + p < p^{2p}$ . In particular, $p^{kp}x^p < p^{2p}$ and so $k\geq 2$ is impossible; also for $k=1$ we have $x<p$ . Now if $x>1$ , then either $x\leq b$ and hence $x$ divides $p^{p}x^{p} - b! = p$ , impossible; or $x>b$ and so $p^pb^p < b! + p < p^{2p}$ , i.e. $b<p$ -- but then in $p^{p}x^p = b! + p$ the term $b!$ would not be divisible by $p$ since $p$ is prime, contradiction.

So it remains to deal with (the most interesting) $p^p = b! + p$ for $p\geq 5$ . Again $p \leq b < 2p$ . Write the equation as $p(p^{p-1} - 1) = b!$ . The right-hand side is divisible by $2^{\frac{p+1}{2}}$ since $b\geq p-1$ and each of $2$ , $4$ , $\ldots$ , $p-1$ is even while for $p\geq 5$ we have the multiplier $4$ which is divisible by $4$ . On the other hand, by the Lifting the exponent lemma for the prime $2$ , the power of $2$ in the left-hand side is $2\nu_2(p-1) + \nu_2(p+1) - 1$ . If $p\equiv 1 \pmod 4$ , then we obtain $2\nu_2(p-1) \geq \frac{p+1}{2}$ , whence $4\log_2 p > p$ and $p^4 > 2^p$ , which fails for $p\geq 17$ by a straightforward induction. If $p\equiv 3 \pmod 4$ , then we obtain $\nu_2(p+1) + 1 \geq \frac{p+1}{2}$ , whence $2\log_2(p+1) \geq p-1$ and $(p+1)^2 \geq 2^{p-1}$ which fails by induction for $p\geq 11$ .

Now we only have to check $p=5, 7, 13$ . Well, $5^5 - 5 = 3120$ is divisible by $13$ but not by $9$ , also $7^7 - 7$ is divisible by $7$ but not by $5$ and $13^{13} - 13$ is divisible by $61$ but not by $19$ .

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megarnie

5610 posts

megarnie

#14 h Jul 13, 2022, 10:12 AM • 3 Y

Y by guywholovesmathandphysics, DEKT, cubres

The answer is $\boxed{(a,b,p)=(2,2,2), (3,4,3)}$ . These work. It remains to show they are the only solutions.

If $a=1$ , then $b!+p=1$ , absurd. So assume $a>1$ .

Claim: $b<2p$ .
Proof: Suppose not. Then $p^2\mid b!$ , which implies $\nu_p(b!+p)=1,$ so it can't be a $p$ th power. $\square$

Case 1: $a=p$ .
Then $b! = p^p-p = p(p^{p-1} - 1)$ If $p=2$ , then $b=2$ , giving $(2,2,2)$

If $p=3$ , then $b=4$ , giving $(3,4,3)$ .

If $p=5$ , then $b!=3120$ , not possible.

If $p=7$ , then $b! = 823536$ , not possible.

Now assume $p>10$ .

By Zsigmondy's, we can find a prime dividing $p^{p-1}-1$ that doesn't divide $p^k -1$ for any $k<p-1$ . So in other words, there exists a prime $q$ , such that the order of $p$ mod $q$ is $p-1$ .

Claim: $q>2p-2$ .
Proof: Assume not.

Since $p\ne q$ , we have $p^{q-1}\equiv 1\pmod q$ , so $p-1\mid q-1$ .

By our assumption that $q\le 2p-2$ , we have $q=p-1$ , which is not prime. $\square$

Now, we have $q\mid p^{p-1}-1\mid p^p - p = b!$ .

Since $q$ is prime, $b\ge q\ge 2p-1$ . Because $b<2p$ , this implies $b = 2p-1$ .

So $p^p - p = (2p-1)!$ .

Claim: $(2n-1)! > n^n$ for any positive integer $n>2$ .
Proof: We have $(2n-1)! = (n-1) ! \cdot n(n+1)(n+2)\cdots (2n-1)$ Notice how $n(n+1)(n+2)\cdots (2n-1)$ is the product of $n$ terms, each of which are at least $n$ .

So $n(n+1)(n+2)\cdots (2n-1)\ge n^n$ , which implies $(2n-1)! \ge (n-1)! n^n > n^n$ , as desired. $\square$

Thus, $(2p-1)! > p^p > p^p - p$ , contradiction.

Case 2: $b<p$ .
Then we have $p\nmid b!$ , so $p\nmid a$ .

Subcase 2.1: $a\le b$ .
Then $a\mid b!$ , so $a\mid a^p - b! = p$ , which implies $a=p$ (since $a>1$ ), absurd.

Subcase 2.2: $a>b$ .
If $p=2$ , then $b=1$ , so $a^2 = 3$ , not possible.

Now assume $p>2$ .

Claim: For any positive integer $n>2$ , we have $2^n > 2n$ .
Proof: We do this by induction on $n$ (base cases $3,4$ ).

If $2^{n-1}> 2(n-1)$ , then $2^n = 2(2^{n-1}) > 4(n-1)>2n,$ as desired. $\square$

So $a^p \ge 2^p >2p$ , which gives $b! > p$ .

Claim: For any positive integer $n>2$ , we have $n^n > 2\cdot n!$ .
Proof: $2\cdot n!$ can be rewritten as $2\cdot 2\cdot 3\cdot 4\cdots n$ . This is a product of $n$ terms, all of which are at most $n$ , and at least one term is less than $n$ .

So $2\cdot n! = 2\cdot 2\cdot 3\cdot 4\cdots n<n^n$ . $\square$

So $a^p > b^p > b^b > 2\cdot b! > b! + p,$ contradiction.

Case 3: $b\ge p$ and $a\ne p$ .
Since $p\mid b!$ , we get $p\mid a$ . Let $a=kp$ with $k>1$ . Then $(kp)^p = b! + p$
Claim: For any positive integer $n>1$ , we have $n^{2n} > (2n-1)! + n$ .
Proof: First note that if $x$ and $y$ are positive, $x,y\ne n$ , and $x+y = 2n$ , then $xy<n^2$ by AM-GM.

So $\begin{align*} (2n-1)! = (1\cdot (2n-1)) \cdot (2\cdot (2n-2)) \cdots ((n-1)\cdot (n+1)) \cdot n\\ < (n^2)^{n-1}\cdot n \\ = n^{2n-1} \\ \le n^{2n}-n \\ \end{align*}$ $\square$

If $k\ge p$ , then $(kp)^p \ge p^{2p} > (2p-1)! + p \ge b! + p$ , contradiction. So $k<p$ .

Let $q$ be a prime divisor of $k$ . Since $q<p\le b$ , we have $q\mid b!$ . So $q\mid (kp)^p - b! = p$ , which implies $q=p$ , contradiction.

We have exhausted all cases, so the solutions described in the beginning are the only ones.

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blackbluecar

304 posts

blackbluecar

#15 h Jul 13, 2022, 1:19 PM • 2 Y

Y by ThisNameIsNotAvailable, cubres

For $p=2,3$ then $(a,b,p)=(2,2,2)$ or $(3,4,3)$ . So, assume $p>3$ .

Rewite as $a^p-p=b!$

We claim $b \geq p$ . Assume $b<p$ . If we let $p_1,p_2,\ldots,p_n$ be the primes $\leq b$ then $p_i$ divides $b!=a^p-p$ for every $1 \leq i \leq n$ . If $p_i$ divides $a$ then $p_i$ divides $(a)^p-(a^p-p)=p$ which is clearly contradiction. Thus, $a \geq p_{n+1}$ . So, $b! = a^p-p \geq p_{n+1}^p-p \geq (p_{n+1})!$ Hence, $b \geq p_{n+1}$ which is contradiction.

Now notice that $p$ divides $b!=a^p-p$ . Thus, $p$ divides $a$ . We claim $b < 2p$ . Indeed, notice that $1=v_p(a^p-p)=v_p(b!)$ So, $b < 2p$ as desired.

Now, we claim $p$ is the smallest prime divisor of $a$ . Indeed, assume some $q<p$ divides $a$ . $q p$ then it must be the case that $a \geq p^2$ . So, $p^{2p}-p \leq a^p-p = b! < (2p)!$ Which cannot hold. So, $a=p$ . plugging back we get $p^p-p=b!$ Now, by LTE we have $v_2(b!)=v_2(p^p-p) = 2v_2(p-1)+v_2(p+1)-1$ Notice that $p^p-p>p!$ so $b \geq p+1$ . Thus, $v_2(b!) \geq v_2((p+1)!) > v_2(p+1)+v_2(p-1)+ v_2 \left ( \frac{p-1}{2} \right ) +1$ which is strictly larger than $2v_2(p-1)+v_2(p+1)-1$ which gives contradiction. $\blacksquare$

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DottedCaculator

7357 posts

DottedCaculator

#16 h Jul 13, 2022, 1:23 PM • 5 Y

Y by asdf334, Mango247, Mango247, Mango247, cubres

Solution

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IAmTheHazard

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IAmTheHazard

#17 h Jul 13, 2022, 3:35 PM • 1 Y

Y by cubres

The answer is $(2,2,2)$ and $(3,4,3)$ which clearly work.
Clearly $b=1$ won't work for size reasons, so assume $b>1$ . From AM-GM, for all $n$ we have $n!<(\tfrac{n+1}{2})^n<n^n$ .
First suppose that $b<p$ , so $p \nmid b!$ and thus $p \nmid a$ . By our bound, we require $a<b$ , else $a^p>a^b\geq b^b>b!$ , but then $a \mid b!-a^p$ and not $p$ : contradiction.
Thus suppose that $b \geq p$ . Since $\nu_p(a^p-p)=1$ , we must also have $b<2p$ . Then we require $a<p^2$ , else $a^p\geq (p^2)^p=p^{2p}\geq (\tfrac{b+1}{2})^{2p}>(\tfrac{b+1}{2})^b>b!$ . Letting $a=kp$ where $k<p$ , it follows that $k \mid a^p-b!$ but not $p$ unless $k=1$ , so we reduce to the case where $a=p$ .
In this case, we count $\nu_2$ : $\nu_2(p^p-p)=\nu_2(p^{p-1}-1)=\nu_2(p^2-1)+\nu_2(\tfrac{p-1}{2})=\nu_2(\tfrac{(p-1)^2(p+1)}{2})\leq \log_2(\tfrac{(p-1)^2(p+1)}{2})$ by exponent lifting, and $\nu_2(b!)\geq b-\log_2(b+1)\geq p-\log_2(p+1)$ by Legendre's. Thus we require
$p-\log_2(p+1)\leq \log_2(\tfrac{(p-1)^2(p+1)}{2}) \implies 2^p \leq \frac{(p-1)^2(p+1)^2}{2}.$ This fails for $p=15$ (yes this isn't a prime, no it doesn't matter), and by derivatives it also fails for all $p>15$ , so we reduce to a finite case check. From this it's easy to get that the solutions provided are the only ones—instead of computing $p^p-p$ , find a divisor of it that can't divide $b$ for each $p$ . $\blacksquare$

The actual expressions are probably wrong but it doesn't really matter

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math90

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math90

#18 h Jul 13, 2022, 4:47 PM • 1 Y

Y by cubres

Answer. $(a,b,p)=(2,2,2)$ and $(a,b,p)=(3,4,3)$ . Those can be easily verified as solutions.

Since $a^p=b!+p>1$ then $a>1$ . Moreover $p\ge 2$ so $a^p\ge 2^p\ge p+2$ . Hence $b\ge 2$ .

If $2\le b<p$ , pick any prime divisor $q$ of $b!+p$ . Hence $q>b$ and $q\mid a$ . It can be proved by induction that $x^y\ge x+y$ for integers $x,y\ge 2$ . Moreover for integers $x,y\ge 2$ we have $xy=(x-1)(y-1)-1+x+y\ge x+y$ so
$b^p=b^{b-1}b^{p-b+1}>b!p\ge b!+p=a^p\ge q^p,$ contradiction.

If $b\ge p$ then $p\mid b!+p=a^p$ so $p\mid a$ . Note that $v_p(a^p-p)=1$ so $b<2p$ . Hence
$\begin{align*} a^p&\le (2p-1)!+p\\ &<(1)(2p-1)(2)(2p-2)\cdots(p-1)(p+1)p+p\\ &\le p[(p^2-1)^{p-1}+1]\\ &\le p^{2p-1}. \end{align*}$ Hence $a<p^2$ . Moreover, if there exists a prime $q<p$ which divides $a$ then $\gcd(q,b!+p)=1$ because $b\ge p$ . But $q\mid b!+p$ , contradiction. Hence $a=p$ so $p^p-p=b!$ .
If $p=2$ we see that $b=2$ so we get the solution $\boxed{(a,b,p)=(2,2,2)}$ . So now assume $p\ge 3$ . By LTE and the fact that $\gcd(p-1,p+1)=2$ we obtain
$\begin{align*} v_2(p^p-p)&=v_2(p^2-1)+v_2\left(\frac{p-1}{2}\right)\\ &=\max\left(2v_2(p-1),v_2(p+1)+1\right)\\ &\le\max\left(2\log_2(p-1),\log_2(p+1)+1\right)\\ &=2\log_2(p-1). \end{align*}$ Moreover
$v_2(b!)\ge v_2(p!)=p-s_2(p)\ge p-\log_2(p+1).$ Hence $2\log_2(p-1)\ge p-\log_2(p+1)$ . Equivalently $(p-1)(p^2-1)\ge 2^p$ . Therefore $2^p<p^3$ so $p<10$ . We are left to check $p=3,5,7$ . If $p=3$ we obtain a solution $\boxed{(a,b,p)=(3,4,3)}$ . If $p=5$ then $p^p-p=5^5-5=3120$ and $6!<3120<7!$ . Hence no solution in this case. If $p=7$ then
$p^p-p=7^7-7\equiv 2^7-2=126\equiv 1\pmod{5}$ Hence $5\nmid 7^7-7$ so for a solution we need $7^7-7\le 4!$ , but $7^7-7>7^2>4!$ . Therefore there is no solution also in this case.

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BOBTHEGR8

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BOBTHEGR8

#20 h Jul 14, 2022, 9:35 AM • 1 Y

Y by cubres

mathisreaI wrote:

Find all triples $(a,b,p)$ of positives integers with $p$ prime and $a^p=b!+p$

Solution-
Claim 1 $b\geq p$
If possible let $b<p$ .
If $a\leq b$ , then $a|b! \implies a|p$ , but $a\leq b<p$ gives $a=1$ which is not possible.
Hence $a>b$ , so $a^p - p = (a-1+1)^p-p > (a-1)^p \geq b^p > b!$ Which is a contradiction, hence $b\geq p$

Claim 2 $a = p$
Going mod $p-1$ , we have $a^p \equiv 1$ , hence order of $a$ is $1$ or $p$ . But order must divide $\phi(p-1)<p$ and hence it is $1$ .
Moreover $b\geq p \implies p|b! \implies p|a \implies a\equiv p \text{ mod} (p^2-p)$ . Hence if $a>p$ then $a\geq p^2$ .
$p^2|a^p \implies p^2$ does not divide $a^p - p = b!$ and hence $b<2p$ .
Now $b!+p \leq (2p-1)! + p < 2 (2p-1)! < 2 \{(2p-1)(1)\}\{(2p-2)(2)\}\cdots \{(p+1)(p-1)\}\{p\} < 2\{p^2\}\{p^2\}\cdots \{p^2\}\{p\} = 2p^{2p-1}< p^{2p} < a^p$ .
Which is a contradiction and hence $a = p$ .

So we have reduced the problem to $b! = p^p-p$ Claim 3 The above equation has no solution for $p>5$
$\nu_2(p^p-p) = \nu_2(p^{p-1}-1) = 2\nu_2(p-1)$ For prime $p\geq 7$ the following are distinct $p-1,\frac{p-1}{2} , 4, 2$ and hence their product divides $b!$
$4(p-1)^2|b! \implies \nu_2(b!) > 2\nu_2(p-1)$ Which is a contradiction.

Now we can check that $p=2,3$ indeed gives two solutions with $b=2,4$ respectively.
Hence all the solution of the given equation are
$(a,b,p) = (2,2,2) \text{ or } (3,4,3)$

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InftyByond

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InftyByond

#63 h Sep 10, 2024, 11:21 PM • 1 Y

Y by cubres

plz help im cooked
can anybody read this solution and tell me how correct (and maybe a score out of 7)

$a^p-a=b!+p-a$ So by FLT we get $a(a^{p-1}-1)\equiv 0 \equiv b!+p-a \mod p.$ Thus, $b!\equiv a \mod p.$ We may then conclude that if $b>p,$ then $b! \equiv 0 \mod p$ .
If $b\leq p$ , then we know that $a^p\geq 2^p \geq 2p \geq b!+p$ which shows that the only solution in this case is $b=a=p=2$ .

So we may now assume that $b! \equiv 0 \mod p$ , and consequently $a \equiv 0 \mod p$ .
Thus $\nu_p(a^p-p)=1$ , and thus $\frac{b!}{p}$ cannot be divisible by $p$ .
Replacing $a$ with $kp$ gives us $a^p-p=k^pp^p-p=p(k^pp^{p-1}-1)=b!$ Dividing both sides by $p$ and taking mod $p$ gives that the right side must be equal to $(b-p)!\equiv 1 \mod p$ .

We show that there are no solutions for $p \geq 5$ .
We first show that $a=p$ .
If $a=kp$ where $k>1$ , then we know that another prime divides $a$ , and if it divides $b!$ then we arrive at a contradiction.
So this means that $k>2p$ .
Then we get $a^p>2^pp^2p>b!+p$ , so $a=p$ .
Then $p^p= (2p-1)!+p$ .
$p^p-p=(2p-1)!$ The right side is clearly greater than the left side for all values $p\geq 5$ , so we only require to test $p=3$ .
Then we get $(3, 4, 3)$ is another solution.

So the only solutions that are possible are $(2, 2, 2)$ , and $(3, 4, 3)$ .
The proof is complete.

This post has been edited 2 times. Last edited by InftyByond, Sep 11, 2024, 12:37 AM

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megarnie

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megarnie

#64 h Nov 18, 2024, 7:55 PM • 1 Y

Y by cubres

Resolving

The only solutions are $(2,2,2)$ and $(3,4,3)$ , which clearly work. Now we show they are the only ones.
Claim: If $p = 2$ , $a = b = 2$ .
Proof: Assume $p = 2$ . We have $a^2 = b! + 2$ . If $b \ge 4$ , then $4\mid b!$ , so $a^2 \equiv 2 \pmod 4$ , a contradiction. Thus, $b \le 3$ and we can check $b \in \{1,2,3\}$ and get that $a = b = 2$ is the only working solution. $\square$

Henceforth assume $p > 2$ .

Claim: $b! > p$ .
Proof: Note that $a = 1$ is obviously not possible. So $b! + p = a^p \ge 2^p > 2p$ , meaning $b! > p$ . $\square$
Claim: $b \ge p$ .
Proof: Suppose not. Since $b! > p > 2$ , $b \ge3$ . Note that $a^p b! + p \le 2b! < (b+1)! < (b+1)^p,$ so $a < b + 1\implies a \le b$ . Thus, $a\mid b!$ . Since $a\mid a^p$ , $a\mid a^p - b! = p$ , so $a \in \{1,p\}$ . Since $a \le b < p$ , $a < p$ , so $a = 1$ , absurd. $\square$

Claim: $p\mid a$ and $b< 2p$
Proof: Firstly, $b \ge p\implies p\mid b! \implies p\mid b! + p = a^p\implies p\mid a$
Now, if $b \ge 2p$ , then $p^2 \mid b!$ , so $\nu_p(b! + p) = 1$ , meaning $\nu_p(a^p) = 1$ , contradiction since $p^p \mid a^p$ . $\square$

Case 1: $a\ne p$
Then $a > p$ . Let $a = xp$ for some $x > 1$ .

Claim: $x \ge p$ .
Proof: Let $q$ be a prime divisor of $x$ . We show that $q \ge p$ , which finishes. Suppose otherwise. Then note that $q\mid b! + p$ but $q \nmid p$ (as $q \ne p$ ), so $q\nmid b! \implies b < q < p$ , absurd. $\square$

We now have $a^p \ge (p^2)^p = p^{2p}$ , so $p^{2p} \le b! + p < (b+1)! \le (2p)!$ . However, note that $(2p)! = p \cdot \prod_{i=1}^{p - 1} i(2p - i) \le p \cdot \prod_{i=1}^{p - 1} p^2 = p^{2p - 1}$ by AM-GM, a contradiction.

Case 2: $a=p$
Then $b! = p^p - p = p(p^{p - 1} - 1)$ . Note that $\nu_2(b!) \ge \nu_2((p)!) = p - s_2(p) \ge p - \log_2(p + 1),$ where $s_2$ is the sum of digits in base $2$ . Now, by LTE $\nu_2(p (p^{p - 1} - 1) ) = \nu_2(p^{p - 1} - 1^{p - 1}) = 2 \nu_2(p - 1) < 2 \log_2(p )$
Combining gives that $p - \log_2(p + 1) < 2 \log_2(p)$ , so $2\log_2(p) + \log_2(p + 1) > p\implies \log_2(p^2(p + 1)) > p$
Thus, $2^p < p^2(p + 1)$ .

Claim: $2^x > x^2(x+1)$ for all integers $x \ge 11$ .
Proof: Let $f(x) = x^2(x+1)$ . We wish to show $2^x > f(x) \forall x \ge 11$ . We go by induction on $x$ . The base case $x = 11$ is obviously true. . Suppose it's true for $11, 12,\ldots, n$ . We have $\frac{f(n+1)}{f(n)} = \left( \frac{n+1}{n} \right) ^2 \cdot \frac{n+2}{n+1} = \left( 1 + \frac 1n \right)^2 \cdot \left( 1 + \frac{1}{n+1} \right) \le \frac{12}{11} ^2 \cdot \left( \frac{13}{12} \right) < 2$
Now, we have $2^{n+1} = 2^n \cdot 2 > f(n) \cdot 2 > f(n) \cdot \frac{f(n+1)}{f(n)} = f(n+1),$ as desired. $\square$

This immediately implies that $p < 11$ , so $p\in \{3,5,7\}$ .

Note that if $p = 7$ , then since $b \ge 7$ , $5\mid b! = 7^7 - 7$ , which is a contradiction as $7^7 \equiv 3 \pmod 5$ .

Now if $p = 5$ , we have $b! = 5^5 - 5 = 3120$ , which isn't a factorial. Therefore, $p = 3$ and $b! = 3^3 - 3 = 24\implies b = 4$ .

Hence $(a,b,p) = (3,4,3)$ , as desired.

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onyqz

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onyqz

#65 h Nov 23, 2024, 12:27 AM • 1 Y

Y by cubres

storage
solution

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EVKV

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EVKV

#66 h Dec 19, 2024, 1:18 PM • 1 Y

Y by cubres

If b≥2p no sol
By mod p then mod p²

If bp then no sol
As a^p≥ (p+1)^p> (p-1)! +p≥b! +p

a≤b
Then a|p
Umpossible

So b<a b! +p

So

b,a≥p

If p= b
And a>b
umpossible

if a p!

(p+1)! > p! +p

So together p^p> p! +p
For p>2
Which gives 1 sol by bounding
2,2,2

If p<a,b<2p

a^p is not divisible by p
But b! +p is
so umpossible

ap has same problem

p<b<2p
a≥2p so a=kp
(kp)^p = b! +p
k=por k≥2p
If k≥2p
At a min
(2p)^p * p^(p) > b! + p

So k= p

If k= p
p^(2p) > b! + p
So only one final case remaining

a= p
p^p = b! +p
Take some odd prime q dividing p-1

2(vq(p-1)) = vq(b!)

If q||p-1
2= vq(b!)>4
So umpossible

q^k||(p-1)
2k = vq(b!) > k + k-1 +k-2 ...... =
(k+1)k /2
2> k+1/2
3>k
Let k= 2
4 = vq(b!)≥ 2+1
vq(b!) = 3 , 5..... So no 4
Umpossible
Thus there is no q

So 2^k = p^(p-1) -1

So only p,2 divide b!

So b≤4

Now done

3,4,3

Final solution (a,b,c) =(3,4,3),(2,2,2)

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cubres

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cubres

#67 h Dec 29, 2024, 10:58 PM

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Storage - grinding IMO problems

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Scilyse

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Scilyse

#68 h Jan 12, 2025, 9:48 AM • 1 Y

Y by cubres

The solutions are $(2, 2, 2)$ and $(3, 4, 3)$ .

If $b \geq 2p$ , then $p \mid b! + p$ , so $p \mid a^p$ , so $p \mid a$ . But $\nu_p(a^p) \geq p \geq 2$ and $\nu_p(b! + p) = \min\{\nu_p(b!), \nu_p(p)\} = 1$ since $2 \leq \nu_p(b!) \neq \nu_p(p) = 1$ , contradiction.

If $p \leq b < 2p$ , then once again $p \mid a$ .

If $a \geq 2p^2$ , then
$\begin{align*} a^p \geq (2p^2)^p = 2^p p^{2p} &= \underbrace{p \cdot p \cdot \dots \cdot p}_{p \text{ times}} \cdot \underbrace{2p \cdot 2p \cdot \dots \cdot 2p}_{p \text{ times}} \\ &> 2 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot \dots \cdot 2p \\ &= 2 \cdot (2p)! \\ &> (2p)! + p \\ &> b! + p, \end{align*}$ contradiction.
If $p^2 \leq a < 2p^2$ , then let for an integer $p \leq k < 2p$ . Now
Therefore, $b! + p > k! + p$ , so $b > k$ , so $k \mid b!$ . Now observe that $k \mid a \mid a^p = b! + p$ , so $k \mid p$ . Since $p \leq k$ , this implies that $k = p$ . Hence, $p^{2p} = b! + p$ .
- If $b \geq p + 1$ , then taking both sides modulo $p + 1$ yields
 $\begin{align*} (-1)^{2p} &\equiv -1 \pmod{p + 1} \\ 1 &\equiv -1 \pmod{p + 1}, \end{align*}$ contradiction as $p + 1 \geq 3$ .
- If $b = p$ , then we claim $p! + p < p^{2p}$ which yields a contradiction. Indeed,
 $\begin{align*} p^{2p} &= \underbrace{p^2 \cdot p^2 \cdot \dots \cdot p^2}_{p \text{ times}} \\ &> 2 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot \dots \cdot p \\ &= 2p! \\ &\geq p! + p, \end{align*}$ as desired.
If $a < p^2$ , then let for an integer so that as above. Now observe that $k \mid a \mid a^p = b! + p$ , so $k \mid p$ . Since $k < p$ , this implies that $k = 1$ . Hence, $p^p - p = b!$ .
- If $p = 2$ , then $(a, b, p) = (2, 2, 2)$ , which works.
- Otherwise, is odd. On one hand, On the other hand, Since is increasing on $[2, \infty)$ , we have Hence $3 \log_2 p - 1 > p - \log_2 p - 1$ , or equivalently, $2^p < p^4$ . This fails for all $p \geq 17$ , so it suffices to check $p \in \{3, 5, 7, 11, 13\}$ .
 - If $p = 3$ , then $(a, b, p) = (3, 4, 3)$ , which works.
 - If $p = 5$ , then $b! = 3120$ , bad.
 - If $p = 7$ , then $b! = 823536$ , bad.
 - If $p = 11$ , then $b! = 285311670600$ , which is contained between $14! = 87178291200$ and $15! = 1307674368000$ , bad.
 - If $p = 13$ , then $b! = 302875106592240$ , which is contained between $16! = 20922789888000$ , $17! = 355687428096000$ , bad.

If $2 \leq b < p$ , then obviously $a < p$ as $p \nmid a$ and if $a \geq p$ then $b! + p \leq \frac{b}{2} \cdot b^{b - 1} + p \leq b^b < p^p \leq a^p,$ contradiction.

If $a < b$ , then $a \mid b!$ and $a \mid a^p \mid b! + p$ , so $a \mid p$ , which implies that $a = 1$ . This obviously yields no solutions.
If $a \geq b$ , then (noting that $a \geq 2$ here)
$\begin{align*} a^p &= (a - 1) \cdot a^{p - 1} + a^{p - 1} \\ &\geq (a - 1) \cdot \underbrace{a \cdot a \cdot \dots \cdot a}_{p - 1 \text{ times}} + p \\ &\geq 1 \cdot 2 \cdot 3 \cdot 4 \cdot \dots \cdot b \cdot 1 \cdot 1 \cdot \dots \cdot 1 + p \\ &= b! + p. \end{align*}$ The condition $a = b = 2$ , derived from the equality case of the second inequality, is necessary for equality to hold throughout. But this implies that $p = 2$ , contradicting the assumption that $b < p$ .

Lastly, if $b = 1$ , then $a^p = p + 1$ . Of course, if $a = 1$ , then $p = 0$ , which is impossible, and if $a \geq 2$ , then we can induct on $p$ to get that $a^p > p + 1$ . Hence, there are no solutions in this case.

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SimplisticFormulas

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SimplisticFormulas

#69 h Jan 25, 2025, 7:50 PM • 2 Y

Y by cubres, radian_51

doesn’t feel too tough for n4

We claim that $(a,b,p)=(2,2,2)$ and $(3,4,3)$ are the only solutions.
Firstly, we show that $b \le 2p$ . Indeed, assume the contrary. Then $p \mid p+ b!=a^p \implies p \mid a \implies p^p \mid a^p$ . But $p^2 \nmid p + b!$ since $p^2 \nmid p$ , giving us the desired contradiction.

Next, we shall prove that $p \mid a$ . Indeed, assume otherwise. Then any prime factor $q$ of a must not $b!$ since $q \mid b!+p$ but $q \nmid p$ . So $b<q \le a$ , But note that $p=a^p - b! >a^p -b^p>p$ , which is absurd.

Now, note that $a \le b$ . If this is not the case then $p=a^p - b! >a^p -b^p>p$ , which is contradictory. This shows that $a=p$ .
Hence, $b!= p^p-p=p(p^{p-1}-1)$ , so $b> \ge p$ From here it’s basically same as $(iii)$ of @Leartia.

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Ilikeminecraft

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Ilikeminecraft

#70 h Mar 13, 2025, 10:33 PM

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If $p = 2,$ we have $a = 2, b = 2.$ If $p = 3,$ $a = 3, b = 3.$ Thus, assume that $p\geq5.$

First, notice that if $b \geq2p,$ then $\nu_p(b!) \geq2,$ while $\nu_p(a^p - p) \leq 1.$ Hence, $b < 2p.$ We also have that if $b = 2p - 1,$ then by Wilsons theorem, $\frac{b! + p}{p} \equiv 2\pmod p,$ and hence can't work.

Now, AFTSOC that $b < p.$ Thus, $a^p - p = b! \leq b^p\implies a \leq b.$ Thus, taking modulo $a,$ we have that $0 \equiv a^p \equiv b! + p \equiv p \pmod a.$ Thus, $a \mid p\implies a = p$ or $a = 1.$ If $a = p,$ then $a \leq b < p = a,$ which is a contradiction.

Thus, $p\leq b < 2p - 1.$ By the upper bound, we have that $b! + p < (2p - 1)! + p < (1\cdot(2p - 1))(2\cdot(2p - 2))\cdot p < (p^2)^p$ By taking modulo $p - 1,$ we have that $a^p \equiv 1 \pmod {p - 1}.$ Thus, $\operatorname{ord}_{p - 1}(a) \mid (\phi(p - 1), p) = 1 \implies a \equiv 1 \pmod {p - 1}.$ Thus, $a \equiv p \pmod{p^2 - p}\implies a = p.$

Thus, we are solving $p^p - p = b!.$ By taking $\nu_2,$ we have that $\nu_2(p^p - p) = \nu_2(p - 1) + \nu_2(p-1) \leq 2\log_2(p -1),$ and $\nu_2(p!) \geq p-\log_2(p+1)$ by legrende. It is obvious that $\nu_2(p!)>\nu_2(p^p-p)$ for $p\geq 10$ from this bound. Manually checking $p = 5, 7$ is trivial.

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ray66

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ray66

#71 h Mar 14, 2025, 4:43 AM

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There are 3 cases.

First if $p>b$ , then $a^p = b! + p$ , so $a$ is not divisible by any prime $\le b$ , and $a \neq 1$ , so the LHS becomes too large because $a^p > a^b > b^p > b!+p$ .

Next if $1 < p \le \frac{b}{2}$ , then $p | a$ because $p$ divides both the RHS and LHS, but the RHS has 1 factor of $p$ because $2p\le b$ , so there are no solutions.

Next if $\frac{b}{2} b$ , the LHS grows too fast. Then $b! = p^p-p$ and $\frac{b!}{p}=p^{p-1}-1$ . LTE gives $\nu_2(b!) = \nu_2(p-1) + \nu_2(p-1) + \nu_2(p+1)-1$ but this means that $b \le 6$ . From here we can check that $(2,2,2)$ and $(3,4,3)$ work

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cursed_tangent1434

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cursed_tangent1434

#72 h Mar 31, 2025, 10:41 AM

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The only solutions are $(2,2,2)$ and $(3,4,3)$ which clearly work. We start off with some bounding.

Claim : We have $p \le b < 2p$ .

Proof : If $b \ge 2p$ note that $p\mid b!+p$ and hence $p\mid a^p$ so $p\mid a$ . But then,
$p^2 \mid a^p - b! = p$ which is a very clear contradiction. Further, if $b<p$ and $a \le b$ then there exists some prime $r<p$ such that $r\mid a$ (clearly $a>1$ ) so
$r \mid a^p - b!=p$ which is again a contradiction. Thus we must have $a>b$ but then,
$a^p \ge (b+1)^p \ge b^p + p\cdot b^{p-1} \ge b^p+p > b!+p$ which is also a contradiction. Thus, there is no possibility for $b$ to violate these bounds, proving the claim.

Claim : We have $a=p$ exactly.

Proof : Since $b \ge p$ we note that,
$p\mid b!+p=a^p$ so $p\mid a$ . Now, say there exists a prime $r<p$ such that $r \mid a$ . But then,
$r \mid a^p-b! =p$ which is a clear contradiction. Thus, such a prime $r$ cannot exist, implying that $a \ge p^2$ , which is impossible since
$a^p\ge p^{2p} \ge (2p)! +p > b! +p$ for all primes $p \ge 3$ (we deal with $p=2$ separately later) or $a=p$ which is the only remaining possibility.

Now, if $p=2$ we have
$a^2=b!+2 \equiv 2 \pmod{4}$ for all $b \ge 4$ which is impossible, implying that $b \in \{1,2,3\}$ of which only $b=2$ works yielding the solution set $(a,b,p)=(2,2,2)$ .

Thus, we are left to sort out the equation,
$p^p-p=b!$ for all primes $p >2$ . Note that if $p=3$ we have $b=4$ yeilding the solution set $(a,b,p)=(3,4,3)$ so we consider $p>3$ in what follows. By the Lifting the Exponent Lemma,
$\nu_2(p^{p-1}-1) = 2\nu_2(p-1)+\nu_2(p+1)-1 \le \max(2\nu_2(p-1)+\nu_2(p+1)+1) \le 2\log_2(p-1)$ However, by Legendre's Formula we have
$\nu_2(b!) \ge \nu_2(p!) \ge \lfloor{\frac{p}{2}}\rfloor$ Further,
$\lfloor{\frac{p}{2}}\rfloor > 2\log_2(p-1)$ for all $p \ge 10$ . Thus, we must have $p \in \{5,7\}$ which are fairly easy to check by hand.

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Maximilian113

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Maximilian113

#73 h Apr 25, 2025, 7:27 PM

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We proceed with casework. Suppose that $a \leq b.$ Then let $q | a$ be a prime, as $a | b!$ it follows that $a^p = b!+p \implies p \equiv 0 \pmod q \implies q=p.$ So $a$ is a power of $p.$ Now note that $b<2p$ as otherwise $v_p$ of the RHS is $1.$ If $a \geq p^2,$ we have that $(2p-1)!+p > b!+p = a^p \geq p^{2p}.$ However by AM-GM $p^{2p}-p \leq (2p-1)! \leq p (1 \cdot (2p-1)) (2 \cdot(2p-2)) \cdots \leq p \cdot \left( p^{p-1}\right)^2 = p^{2p-1},$ which is a contradiction. So $a=p.$ We deal with this case later.

If $a > b,$ note that $b!+p > b^p \implies b \geq p.$ Hence $b < 2p$ by the same argument as above. So $p | (b!+p) \implies p|a.$ By the same argument as above $a < p^2.$ But then write $a=kp$ with $k < p.$ If $k \neq 1$ taking mod $k$ yields $0 \equiv b! + p \equiv p \pmod k \implies k | p \implies k = p,$ a contradiction. So $k=1 \implies a=p,$ though this is impossible as $p=a>b\geq p.$

Now it suffices to deal with the case that $a=p.$ The equation becomes $b!=p^p-p.$ From above $2p > b \geq p.$ Suppose that $p \equiv 1 \pmod 4.$ Then by LTE $v_2(b!) = v_2(p^{p-1}-1) = 2v_2(p-1)+v_2(p+1)-1 =2v_2(p-1).$ If $p \geq 13,$ observe that $\{2, 4, (p-1)/2, p-1\}$ is a set of distinct elements, and each of these are in the product of $b!$ as $b \geq p.$ Then $v_2(b!) \geq v_2(4)+2v_2(p-1),$ a contradiction. So it suffices to check $p=5,$ but this clearly does not work.

Now if $p \equiv 3 \pmod 4,$ by LTE $v_2(b!) = 2v_2(p-1)+v_2(p+1)-1 = v_2(p+1)+1.$ But for $p \geq 11$ $\{2, 4, (p+1)/2\}$ is a set of distinct elements in the product of $b!,$ as $b \geq p$ so $v_2(b!) \geq v_2(p+1)+2,$ contradiction. So we should manually check $p=3, 7.$ $p=7$ does not work as $5$ does not divide $7^7-7$ so $b \leq 4,$ contradiction. $p=3$ works. Finally if $p$ is even $p=2$ and this works. Hence the only solutions are $(a, b, p) = (3, 4, 3), (2, 2, 2).$ They clearly work.

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Jlzh25

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Jlzh25

#76 h May 17, 2025, 2:48 PM

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mathisreaI wrote:

Find all triples $(a,b,p)$ of positive integers with $p$ prime and $a^p=b!+p.$

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Adywastaken

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Adywastaken

#78 h May 20, 2025, 4:10 PM

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Assume FTSOC $b<p$ .
Case 1: $a\le b<p$
$a\mid p \implies a=1$ , but $1\neq b!+p$

Case 2: $a>b$
$a^p-p>(b+1)^p-p>b^p>b!$
So, $b\ge p$ . Then, $a=kp$ and $v_p(b!)=v_p((kp)^p-p)=1$ , so $b\le 2p$ .

If $a\ge p^2$ , then $b!+p=a^p>p^{2p}$ , $\Rightarrow \Leftarrow$ .

Now, $b!$ and $a$ have no common factors apart from $p$ , so $k=1$ .

$p^p-p=b!$ , so
$v_2(p^p-p)=v_2(p^{p-1}-1)=2v_2(p-1)=v_2(b!).$ So, if $p-1$ , $\frac{p-1}{2}$ , $4$ all divide $b!$ , $\Rightarrow \Leftarrow$

Thus, $p=5$ or $p<5$ .

Checking gives $(a,b,p) = (2,2,2), (3,4,3)$ .

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maromex

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maromex

#79 h May 20, 2025, 4:51 PM

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Wow I just solved this problem 40 mins after @above lol
Take mod $p$ , so we find that $a \equiv b! \pmod p$ . So $p$ divides $a$ if and only if $b \ge p$ .
By the way, what do we think is the answer?
For example, $(2,2,2),(3, 4, 3)$ .
Consider the case where $b \ge p$ . After taking mod $p^2$ , then we must also have $b \le 2p$ . We will prove that $a < p^2$ . We must first manually prove this for $p = 2$ . If $a = 4$ , then $14 = b!$ , contradiction. If $a = 5$ then $23 = b!$ , contradiction. If $a \ge 6$ then we see that $24 = 4! \ge b! > 34$ which is a contradiction.

Now, suppose $p \ge 3$ . Assume for contradiction that $a \ge p^2$ . Then we must have $b! + p \ge p^{2p}.$ Because $b \le 2p$ , we must have $(2p)! + p \ge p^{2p}.$ However, we can prove that $(2p)! + p < p^{2p}$ by induction on $p$ , as a positive integer. The statement holds for $p = 3$ , because $723 < 729$ . Now assume the statement holds for $p=k$ . Then, we need to prove $(2k + 2)! + k + 1 < (k + 1)^{2k+2}.$ We have $(2k)! < k^{2k} - k$ , which implies $(2k + 2)! + k + 1 < (4k^2 + 6k + 2)(k^{2k} - k) + k + 1.$ Notice that, by the Binomial Theorem, $(k + 1)^{2k+2} \ge k^{2k+2} + (2k + 2)k^{2k+1} + (k+1)(2k+1)k^{2k} + \frac{1}{3}(k+1)(2k+1)(2k)k^{2k-1}$ $=\left(\frac{19}{3}k^2 + 7k + 1\right)k^{2k} > (4k^2 + 6k + 2)k^{2k}.$ From here our goal follows, and the induction case is done.

A prime less than $p$ cannot divide $b! + p$ , because it divides $b!$ but not $p$ . Thus, such a prime cannot divide $a^p$ either. Therefore, $a = p.$
So we have $p^p = b! + p \implies p(p^{p-1} - 1) = b!.$ If $p = 2$ , only solution is $(2, 2, 2)$ . Using Lifting the Exponent lemma, for all odd primes $p$ , $\nu_2(p^{p-1} - 1) = 2\nu_2(p-1) + \nu_2(p+1) - 1.$ We have already found the working triple for $p = 3$ . If $p = 5$ , then we get $3120 = b!$ which is never true.

If $p = 7$ , then we see that $43 \mid 344 \mid (7^3 -+ 1)(7^3 - 1) \mid 7(7^6 - 1) = b!$ , which is a contradiction because $b \le 14$ but $b \ge 43$ .

If $p = 11$ , Legendre's formula tells us that $\nu_2(b!) \ge \nu_2(11!) = 8$ . However, we know that $\nu_2(p(p^{p-1} - 1)) = 3$ , contradiction. If $p = 13$ , we see that $\nu_2(b!) \ge \nu_2(13!) = 10$ , but $\nu_2(p(p^{p-1} - 1)) = 4$ . If $p = 17$ , we get $\nu_2(b!) \ge \nu_2(17!) = 15$ , but $\nu_2(p(p^{p-1} - 1)) = 8$ .

If $p \ge 19$ , notice that one of $\nu_2(p-1)$ and $\nu_2(p+1)$ is equal to $1$ , and both of them are at most $\log_2(p)$ . Therefore, $\nu_2(p^{p-1} - 1) \le 2\log_2(p)$ . Also, $\nu_2(b!) \ge \frac{p-1}{2}$ . However, we find that $2 \log_2(p) < \frac{p-1}{2}.$ This is a contradiction. Therefore, the only solutions in this case are $(2, 2, 2)$ and $(3, 4, 3)$ .

The next case is where $b < p$ . We have $a \neq 1$ , because $b! + p > 1$ . Now, $a^p > (a - 1)! + p$ for all positive integers $p \ge 2$ and $a \ge 2$ , which can be shown by induction on $p$ . Thus, the given equation $a^p = b! + p$ cannot be satisfied for $a > b$ . We must have $a \le b$ . Now, taking the given equation mod $a$ , we get $0 \equiv p \pmod a.$ Therefore, $a$ divides $p$ . Because $a$ is not $1$ , we must have $a = p$ . But then $b \ge p$ , contradiction. Indeed the only solutions are $(2, 2, 2), (3, 4, 3)$ .

This post has been edited 2 times. Last edited by maromex, May 20, 2025, 4:53 PM
Reason: fix

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lolsamo

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lolsamo

#80 h May 20, 2025, 5:07 PM

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The post name should be Belgium number theory, lol.

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