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Source: 2023 ISL N2

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#1 h Jul 17, 2024, 12:01 PM • 5 Y

Y by OronSH, peace09, MarkBcc168, BorivojeGuzic123, Efesc128e968

Determine all ordered pairs $(a,p)$ of positive integers, with $p$ prime, such that $p^a+a^4$ is a perfect square.

Proposed by Tahjib Hossain Khan, Bangladesh

This post has been edited 1 time. Last edited by avisioner, Jul 20, 2024, 4:57 PM
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#2 h Jul 17, 2024, 12:02 PM • 4 Y

Y by ehuseyinyigit, Tellocan, poirasss, Efesc128e968

$p^a+a^4=k^2\iff p^a=(k-a^2)(k+a^2)$ $i)p\geq 5,$
$k-a^2=p^x$ and $k+a^2=p^{a-x}$
$2a^2=p^{a-x}-p^x=p^x(p^{a-2x}-1)$ $x=v_p(RHS)=2v_p(a). \$ Let $a=p^{\frac{x}{2}}b$ where $(p,b)=1$ .
$2b^2+1=p^{a-2x}=p^{p^{\frac{x}{2}}b-2x}$ But
$2b^2+1=p^{p^{\frac{x}{2}}b-2x}\geq p^{5^{\frac{x}{2}}b-2x}\geq p^b\geq 5^b>2b^2+1$ Since $p\geq 5, \ 2b^2+1<5^b$ and $(5^{\frac{x}{2}}-1)b\geq 5^{\frac{x}{2}}-1\geq 2x$ which can be proved by induction. $\square$
$ii)p=3$
$3^a=(k-a^2)(k+a^2)$ $k-a^2=3^x$ and $k+a^2=3^{a-x}$
$2a^2=3^x(3^{a-2x}-1)$ Denote $a=3^{\frac{x}{2}}b$ We have
$2b^2+1=3^{3^{\frac{x}{2}}b-2x}$ If $b=1,$ then $3^{\frac{x}{2}}-2x=1$ which gives $x=0,4$ so we get $\boxed{(a,p)=(9,3),(1,3)}$ . If $b=2,$ then $2.3^{\frac{x}{2}}-2x=2\iff 3^{\frac{x}{2}}-x=1\implies x=0,2$ Thus we get $\boxed{(a,p)=(6,3),(2,3)}$
Suppose that $b\geq 3$ .
$3^b> b^2+1=3^{3^{\frac{x}{2}}b-2x}\implies 2x>b(3^{\frac{x}{2}}-1)\geq 3(3^{\frac{x}{2}}-1)\geq 6x$ Which is impossible. $\square$
$iii)p=2$
$k-a^2\neq 1$ Let $k-a^2=2^x, \ k+a^2=2^{a-x}$
$2a^2=2^x(2^{a-2x}-1\iff a^2=2^{x-1}(2^{a-2x}-1)$ Let $x=2m+1$ and $a=2^mb$
$b^2+1=2^{2^mb-4m-2}$ For $m=1$ we have $b^2+1=2^{2b-6}\underbrace{>}{\text{for} \ b\geq 6}b^2+1$ which gives no solution and for $m=2$ we have $b^2+1=2^{4b-10}\underbrace{>}{\text{for} \ b\geq 4}b^2+1$ which gives no solution again. If $m\geq 3,$ then $2^m>4m+2$ hence
$2^{b+1}\geq b^2+1=2^{2^mb-4m-2}>2^{(4m+2)(b-1)}\geq 2^{12(b-1)}\implies b+1\geq 12(b-1)\implies b=1$ Which gives a contradiction. $\blacksquare$

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OronSH

#3 h Jul 17, 2024, 12:05 PM • 1 Y

Y by peace09

Answer: $p=3,a=1,2,6,9.$

Set $p^a+a^4=b^2,$ then $p^a=(b-a^2)(b+a^2).$ Then set $b-a^2=p^x,b+a^2=p^y$ where $x,y$ are nonnegative integers with $y>x$ and $x+y=a.$ We also get $2a^2=p^y-p^x$ so we are trying to solve $p^y-p^x=2(x+y)^2.$

First suppose $p=2.$ We get $2^{y-x}-1=\frac{(x+y)^2}{2^{x-1}}.$ Then $\nu_2$ gives $x-1$ is even, so $\frac{(x+y)^2}{2^{x-1}}$ is a square and an integer. Thus we have $y-x=1$ by Mihailescu, so plugging this in we have $2^{x-1}=(2x+1)^2.$ Now the right side is odd, so $x=1$ which doesn't work. Thus there are no solutions in this case.

Now if $p\ne 2,$ we have $p^x\cdot\frac{p^{y-x}-1}2=(x+y)^2.$ By $\nu_p$ we have $x$ even, so $p^x$ and $\frac{p^{y-x}-1}2$ are both squares. However, $\frac{p^{y-x}-1}2\equiv-\frac12\pmod p.$ Now it is well known that $-1$ is a QR $\pmod p$ iff $p\equiv 1\pmod 4,$ and $2$ is a QR $\pmod p$ iff $p\equiv 1,7\pmod 8.$ Thus $-\frac12$ is a QR iff $p\equiv 1,3\pmod 8.$ Thus $p\ne 5,7.$

Now if $p\ge 11,$ we have $8y^2=2(y+y)^2>2(x+y)^2=p^y-p^x\ge p^y-p^{y-1}\ge10p^{y-1}>8p^{y-1},$ so $y^2>p^{y-1}.$ But for $y=1$ we have $y^2=p^{y-1},$ and $p^{y-1}$ grows faster than $y^2$ since $\frac{p^y}{p^{y-1}}\ge 11$ but $\frac{(y+1)^2}{y^2}\le\frac{(y+y)^2}{y^2}=4$ for $y\ge 1.$ Thus there are no solutions in this case.

Now we consider $p=3.$ We have $8y^2>2(x+y)^2=3^y-3^x\ge 3^y-3^{y-1}=2\cdot 3^{y-1},$ so $4y^2>3^{y-1}.$ This is true for $y=1,2,3,4,5$ but false for $y=6,$ and thus false for $y\ge 6$ since $\frac{(y+1)^2}{y^2}=\frac{y^2+2y+1}{y^2}<\frac{y^2+y^2+y^2}{y^2}=3$ for $y\ge 6,$ so the right side grows faster than the left.

Thus we just have to check $y=1,2,3,4,5$ and $x<y.$ Checking, we get only $(y,x)=(1,0),(2,0),(4,2),(5,4)$ work. Using $a=x+y$ we get $a=1,2,6,9.$ These work, so we are done.

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#4 h Jul 17, 2024, 12:14 PM • 1 Y

Y by ehuseyinyigit

Time consuming, but great to exercise optimization strategies. Answer. $(p,a)$ : $(3,1), (3,2)$ , $(3,6)$ , $(3,9)$ .

The equation $p^a + a^4 = m^2$ is equivalent to $p^a = (m - a^2)(m + a^2)$ , hence (as $p$ is prime) $m-a^2 = p^x$ and $m + a^2 = p^{a-x}$ for some $0 \leq x < \frac{a}{2}$ . Thus $2a^2 = p^{a-x} - p^x$ . In the right-hand side the first time decreases as $x$ increases, while the second term increases, hence

$2a^2 \geq p^{a - \left \lfloor \frac{a-1}{2} \right\rfloor} - p^{\left \lfloor \frac{a-1}{2} \right\rfloor} = p^{\left \lfloor \frac{a-1}{2} \right\rfloor}(p^{a - 2\left \lfloor \frac{a-1}{2} \right\rfloor} - 1).$
If $a=1$ , then in $2a^2 = p^{a-x} - p^x$ we can have only $x=0$ , so $p=3$ . If $a=2$ , then again $x=0$ and $8 = p^2 - 1$ yield $p=3$ .

Suppose $a\geq 3$ and $p\geq 5$ . In the above inequality the right-hand side is at least $p^{\left \lfloor \frac{a-1}{2} \right\rfloor}(p-1) \geq 5^{\left \lfloor \frac{a-1}{2} \right\rfloor} \cdot 4$ , so necessarily $a^2 \geq 2 \cdot 5^{\frac{a-1}{2}}$ . This is false for $a=3$ (direct check) and for $a\geq 5$ by induction: the base case is checked directly, and if $5^{\frac{a-1}{2}} > \frac{a^2}{2}$ , then multiplying with $5 > \frac{(a+1)^2}{a^2}$ (the right-hand side is indeed not more than $\left(1 + \frac{1}{a}\right)^2 \leq \left(\frac{6}{5}\right)^2 < 2$ ) we obtain what we want. If $a=4$ in $2a^2 = p^{a-x} - p^x$ we have $2^5 = p^{4-x} - p^x$ and $0\leq x < 2$ , but none of $x=0$ and $x=1$ yields a solution (in the first case $p^4 = 33$ , in the second one $p$ divides the right-hand side, so $p=2$ which does not work).

Now let $p=3$ and $a\geq 3$ . In $2a^2 = p^{a-x} - p^x$ if $x=0$ , then $2a^2 = 3^a - 1$ , impossible since the right-hand side is larger by induction. If $x\geq 1$ , then $3$ divides $2a^2$ and so $a$ is divisible by $3$ . For $a=3$ the main expression is $3^3 + 3^4 = 108$ , for $a=6$ is $3^6 + 6^4 = 45^2$ , for $a=9$ it is $3^9 + 9^4 = 3^9 + 3^8 = 3^8 \cdot 4 = 162^2$ . For $a\geq 12$ in the main inequality the right-hand side is at least $p^{\left \lfloor \frac{a-1}{2} \right\rfloor}(p-1) \geq 2 \cdot 3^{\frac{a-1}{2}}$ , but actually $3^{\frac{a-1}{2}} > a^2$ by induction for $a\geq 11$ (for $a=11$ we have $243 > 121$ and the step follows by $3 > \left(1+\frac{1}{a}\right)^2$ ).

It remains to consider $p=2$ and $a\geq 3$ . In the main inequality the right-hand side is at least $2^{\frac{a-1}{2}}$ , so $a^2 \geq 2^{\frac{a-3}{2}}$ , false for $a\geq 21$ (by induction, since $512 > 441$ and $2 > \left(1 + \frac{1}{a}\right)^2$ ). For $a=1,2,\ldots,20$ the expression $2^a + a^4$ equals $3 \in (1^2, 2^2)$ , $20 \in (4^2, 5^2)$ , $89 \in (9^2, 10^2)$ , $272 \in (16^2, 17^2)$ , $657 \in (25^2, 26^2)$ , $1360 \in (36^2, 37^2)$ , $2529 \in (50^2, 51^2)$ , $2^8 + 2^{12} = 2^8 \cdot 17$ , $7073 \in (84^2, 85^2)$ , $11024 \in (104^2, 105^2)$ , $16689 \in (129^2, 130^2)$ , $2^{12} + 12^4 = 2^{8}(16 + 81) = 2^8 \cdot 97$ , $2^{13} + 13^4$ ends at $3$ , impossible for a square; $2^{14} + 14^4 = 2^4(1024 + 2401) = 2^4 \cdot 5^2 \cdot 137$ and $137 \in (11^2, 12^2)$ is not a square; $2^{15} + 15^4$ ends at $3$ , impossible for a square; $2^{16} + 16^4 = 2^{17}$ and is not a square; $2^{17} + 17^4$ ends at $3$ , impossible for a square; $2^{18} + 18^4 = 2^4 \cdot 22945$ , the second multiplier is divisible by $5$ , but not by $25$ ; $2^{19} + 19^4 = 654609$ is divisible by $3$ , but not by $9$ ; $2^{20} + 20^4 = 2^8(2^{12} + 5^4) = 2^8 \cdot 4721$ , where $4721 \in (68^2, 69^2)$ is not a square. Therefore there is no solution for $a=2$ .

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#5 h Jul 17, 2024, 1:30 PM • 1 Y

Y by isomoBela

Rewrite $p^a+a^4=x^2$ by factoring out $p^a = (x-a^2)(x+a^2)$ . The second bracket is strictly positive, so the first one must be as well. We now consider several cases for $x - a^2$ :

$x - a^2 = 1$ . Then $x = a^2+1\Longrightarrow p^a = 2a^2 + 1$ . The right-hand side is odd, so $p \geq 3$ . Now $p^a \geq 3^a > 2a^2 + 1$ for $a>2$ (by induction, or taking the derivative of $f(a) = 3^a - 2a^2 - 1$ , etc.), and $a=1, 2$ lead to the solutions $(a,p) = (1,3), (2,3)$ .
$x - a^2 > 1$ and $p > 2$ . Then $p\mid x-a^2$ and $p\mid x+a^2$ , so $p\mid 2x$ and $p\mid 2a^2$ . As $p>2$ , these imply $p\mid x$ and $p\mid a$ . Now if $\nu_p(x) = X$ and $\nu_p(a) = A$ , and $x = x_1p^{X}$ , $a = a_1p^{A}$ , then we can write the initial equation as: $p^{a_1p^{A}} + a_1^4 p^{4A} = x_1^2 p^{2X}.$ As $\nu_p(x^2) = 2X$ is even, so must be $\nu_p(p^{a_1p^A}+a_1^4p^{4A}$ . However, if we assume that we don't just have $a_1p^{A} > 4A$ , then $a_1 = 1$ , $p = 3$ , and $A = 1$ , at which point $\nu_p(p^a+a^4 = 3$ , which is odd, contradiction. Therefore, $a_1p^A > 4A$ and so
$4A = \nu_p(p^a+a^4) = \nu_p(x^2) = 2X \Longrightarrow X = 2A.$ Dividing both sides by $p^{4A}$ leaves $p^{a_1p^{A} - 4A} = (x_1 - a_1^2)(x_1 + a_1^2)$ . Now if $p\mid x_1-a_1^2$ , then $p\mid x_1+a_1^2$ as well, so $p\mid a_1$ and $p\mid x_1$ , contradiction with the assumption that $\nu_p(x) = X$ and $\nu_p(a) = A$ from before. Hence $x_1 - a_1^2 = 1$ and so
$p^{a_1p^{A} - 4A} = 2a_1^2+1.$ If $a_1>2$ , then $p^{a_1p^{A} - 4A} \geq p^{3a_1-4} \geq 3^{3a_1-4} \geq 2a_1^2+1$ . Checking $a_1 = 1, 2$ leads to the solutions $(a,p) = (9,3)$ and $(a,p) = (6,3)$ , respectively.
$x-a^2 > 1$ and $p = 2$ . Here $x-a^2 = 2^y$ and $x+a^2 = 2^z$ for some positive integers $y<z$ , and so $2^z - 2^y = 2a^2\Longrightarrow 2^{z-1} - 2^{y-1} = a^2$ . Diving by $4$ until we can, we must get to $2^{z-y} - 1 = \alpha^2$ for some $\alpha \in\mathbb{N}$ . However, $2^{z-y} - 1 \equiv 3\not\equiv \alpha^2 \pmod 4$ for $z>y+1$ , so we must have $z = y+1$ . This leads to $x+a^2 = 2(x-a^2)$ , or $x = 3a^2$ . Plugging this into the original equation gives $2^{a} + a^4 = 9a^4$ , so $2^a = 8a^4$ . At this point, $a$ is some power of two, say $a = 2^b$ , but $2^a = 8a^4 \Longrightarrow 2^b = 4b+3$ , which is impossible due to modulo $4$ issues again. Thus, there are no solutions in this case, and we're done.

In conclusion, all solutions are $(a,p) = (1,3), (2,3), (6,3), (9,3)$ .

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#6 h Jul 17, 2024, 1:40 PM

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Let $(a, p)$ be a pair of integers satisfying the equation $p^a + a^4 = k^2$ for some integer $k$ . Then we have:
$p^a = (k - a^2)(k + a^2) = p^m \cdot p^{a-m}$ where $k - a^2 = p^m$ , $k + a^2 = p^{a-m}$ , and $m$ is a non-negative integer.

Subtracting these two equations gives:
$p^{a-m} - p^m = 2a^2$
This simplifies further to:
$p^m(p^{a-2m} - 1) = 2a^2$
It is clear that $p \neq 2$ and $p \nmid \frac{p^{a-2m} - 1}{2} = \frac{a^2}{p^m}$ , leading to $v_p(a^2) - v_p(p^m) = 0$ , implying $2v_p(a) = m$ .

Therefore, $m = 2b$ and $a = c \cdot p^b$ for some positive integer $c$ and non-negative integer $b$ . Hence,
$\frac{p^{a-2m} - 1}{2} = c^2 = \frac{p^{c \cdot p^b - 4b} - 1}{2}$
Since $p \neq 2$ , we have $p \geqslant 3$ . Considering $b$ as an integer, the inequality $3^b - 1 \geqslant 4b - 2$ is always true. Furthermore, since $p \geqslant 3$ , we have $p^b - 1 \geqslant 3^b - 1 \geqslant 4b - 2$ , which results in:
$c \geqslant 1 \geqslant \frac{4b - 1}{p^b - 1} \implies cp^b - c \geqslant 4b - 1 \implies cp^b - 4b \geqslant c - 1$
Next, using the increasing nature of the function $f(x) = p^x$ , we obtain:
$\frac{1}{2}(p^{cp^b - 4b} - 1) \geqslant \frac{1}{2}(p^{c - 1} - 1) \implies c^2 \geqslant \frac{1}{2}(p^{c - 1} - 1) \geqslant \frac{1}{2}(3^{c - 1} - 1)$
Upon further observation, if $c \geqslant 5$ , the inequality $c^2 < \frac{1}{2}(3^{c - 1} - 1)$ holds true.

In conclusion, the possible values for $c$ are in the range $1 \leqslant c \leqslant 4$ . Let's consider each case:
$\bullet$ If $c = 1$ , then $2 = p^{p^b - 4b} - 1$ , which implies $p = 3$ and $p^b - 4b = 1$ . Thus, $b = 2$ or $b = 0$ . If $b = 2$ , then $a = 9$ , satisfying $p^a + a^4 = 3^9 + 9^4 = 162^2$ , a perfect square. If $b = 0$ , then $a = 1$ , satisfying $p^a + a^4 = 3^1 + 1^4 = 2^2$ , also a perfect square.

$\bullet$ If $c = 2$ , then $8 = p^{2p^b - 4b} - 1$ , implying $p = 3$ and $2p^b - 4b = 2$ . Thus, $b = 0$ or $b = 1$ . If $b = 0$ , then $a = 2$ , satisfying $p^a + a^4 = 3^2 + 2^4 = 5^2$ , a perfect square. If $b = 1$ , then $a = 6$ , satisfying $p^a + a^4 = 3^6 + 6^4 = 45^2$ , also a perfect square.

$\bullet$ If $c = 3$ , then $18 = p^{3p^b - 4b} - 1$ leads to no integer solutions.

$\bullet$ If $c = 4$ , then $32 = p^{4p^b - 4b} - 1$ results in $33 = p^{4p^b - 4b}$ , but $33$ is not a power of any prime number, hence no solutions exist in this case.

Therefore, the valid pairs $(a, p)$ that satisfy all conditions are $\boxed{(1, 3), (2, 3), (6, 3), (9, 3)}$ . Each pair has been verified to meet all criteria.

This post has been edited 1 time. Last edited by EeEeRUT, Jul 17, 2024, 1:43 PM
Reason: ITEMIZE

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#7 h Jul 17, 2024, 3:01 PM • 6 Y

Y by OronSH, vsamc, Kawhi2, alexanderhamilton124, TensorGuy666, eezad3

the only correctly placed problem on the n shortlist is hard for the wrong reasons

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#8 h Jul 17, 2024, 4:22 PM • 5 Y

Y by avisioner, MS_Kekas, sami1618, ihatemath123, alexanderhamilton124

This problem was proposed by Tahjib Hossain Khan, Bangladesh.

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dkedu

#9 h Jul 22, 2024, 8:11 AM

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We claim the only solutions are $(1,3), (2,3), (6,3), (9,3)$ .

Let $p^a + a^4 = x^2$ so we have that
$p^a = (x-a^2)(x+a^2) \implies x-a^2 = p^k, x+a^2 = p^{a-k} \implies a^2 = \frac{p^{a-k} - p^k}2.$
If $a = 1$ , we get the only solution is $(1,3)$ . We will proceed with casework.

Case 1: $p \ge 5$ .
In this case, we get that $\nu_p(x) = 2\nu_p(a) = 2r$ where $2r \ge k-1$ . If $r = 0$ , we have $a^2 \ge \frac{p^{a-1} - p}{2} \ge 2p^{a-2}$ unless $a = 1$ or $a=2$ which we verify to not work. We have that $\frac{p^{a-k}-p^k}{2} \ge 2p^{a-k-1} \ge \frac{2p^{a-2}}{a^2} > a^2$ where the last inequality holds unless $a = 5$ and $p = 5$ , but the remaining case can be checked to fail. So there are no solutions here.

Case 2: $p = 3$ .
Here, we have that $\nu_p(x) = 2\nu_p(a) = 2r, 2r\ge k - 1$ unless $r=1$ and $a = 3$ but this does not result in a solution. So we get that $a^2 = \frac{p^{a-k}-p^k}{2} \ge p^{a-k-1} \ge \frac{p^{a-2}}{a^2}.$ This reduces to a case check for $a < 10$ which results in solutions $(2,3), (6,3), (9,3)$ .

Case 3: $p = 2$ .
Here, we have that if $k = 0$ , then $2^a = 2a^2 + 1$ which clearly has no solutions. So we can get that $a^2 = 2^{k-1}(2^{a-2k}-1)$ so $k$ is odd. Let $k = 2l + 1$ and $a = 2^lq$ , we get that $q^2 \equiv 3\pmod 4$ unless $a = 2k$ or $a = 2k+1$ . But these cases give no solution as the first implies $a = 0$ and the second means $k = 1$ and this does not give a solution.

Having exhausted all cases, we can conclude that the solutions we initially listed are the only ones.

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#11 h Jul 26, 2024, 2:48 PM

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The pairs are $\boxed{(1,3),(2,3),(6,3),(9,3)}$ .

Let $k$ be a positive integer such that $p^a+a^4=(a^2+k)^2\iff p^a=k(2a^2+k)$
We quickly resolve the case $p=2$ . We can set $k=2^{k'}$ giving $2^{a-k'}=2a^2+2^{k'}$ . We thus must have $2a^2=2^{k'}$ or $a=2^{\frac{k'-1}{2}}$ . Thus we must have $2^{\frac{k'-1}{2}}=2k'+1$ . Checking $k'\leq 10$ gives no solutions and one can finish with induction showing the LHS is larger.

Now set $k=p^{2k'}$ and we must have $a=a'p^{k'}$ for an non-negative $k'$ . The case $a'=k'=1$ clearly gives no solutions. Then we must have $p^{a'p^{k'}-4k'}=2{a'}^2+1$ Notice that for $(a',k')\neq (1,1)$ we have $a'p^{k'}-4k'\geq a'3^{k'}-4k'\geq a'$ . Thus we must have $3^{a'}\leq p^{a'}\leq 2{a'}^2+1$ . Thus we must have either $a'=1$ or $a'=2$ and $p=3$ .

If $a'=1$ the equality simplifies to $3^{3^{k'}-4k'}=3$ so we must have $3^{k'}=4k'+1$ which has solutions at $k'=0$ and $k'=2$ giving $(a,p)$ as $(1,3)$ and $(9,3)$ .

If $a'=2$ the equality simplifies to $3^{2\cdot 3^{k'}-4k'}=9$ so we must have $3^{k'}=2k'+1$ which has solutions at $k'=0$ and $k'=1$ giving $(a,p)$ as $(2,3)$ and $(6,3)$ .

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alexanderhamilton124

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alexanderhamilton124

#12 h Sep 19, 2024, 10:36 AM

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Let $p^a + a^4 = k^2$ . We have $(k - a^2)(k + a^2) = p^a$ > If $k - a^2 = 1$ , then $2a^2 + 1 = p^a$ . We have that $p^a \geq 2^a > 2a^2 + 1$ , for $a = 7$ . Say $2^a > 2a^2 + 1$ for $a = \{7, \dots, k\}$ . Then, $2^k > 2k^2 + 1$ , and $2^{k + 1} = 2 \cdot 2^k > 4k^2 + 2 > 2(k + 1)^2 + 1$ , so $2^a > 2a^2 + 1$ for $a \geq 7$ , now we simply have to check $a = 1, 2, \dots 6$ , which yield $(1, 3), (2, 3)$ as solutions.

Now, assume $k - a^2 \neq 1$ . Then $p \mid k - a^2, p \mid k + a^2 \implies p \mid a, k$ . Let $k = p^{\alpha}x$ and $a^2 = p^{2\beta}y$ , where $p \nmid x, y$ . Let:
$k - a^2 = p^{\gamma}, \quad k + a^2 = p^{a - \gamma}$ First, note that $a > 2\gamma$ . We have $p^{\alpha}x + p^{2\beta}y = p^{a - \gamma}$ , and WLOG $\alpha \geq 2\beta$ which gives $p^{2\beta}(p^{\alpha - 2\beta}x + y) = p^{a - \gamma}$ . If $\alpha \neq 2\beta$ , then $p \nmid p^{\alpha - 2\beta}x + y$ since $p \nmid y$ , a contradiction (as it's not equal to 1). Therefore $v_p(k) = v_p(a^2)$ , and $\alpha = 2\beta$ .

So, we have $p^{2\beta}(x - y) = p^{\gamma}, p^{2\beta}(x + y) = p^{a - \gamma}$ . Note that $x + y \neq 1$ , so $p \mid x + y$ . If $x - y \neq 1$ , then $p \mid x - y$ as well which yields $p \mid x, y$ a contradiction therefore $x - y = 1$ , resulting in $p^{2\beta} = p^{\gamma}$ , and $2y + 1 = p^{a - 2\gamma}$ , and since $2y + 1 = \frac{2a^2}{p^{\gamma}} + 1$ , we have $2a^2 = p^{a - \gamma} - p^{\gamma}$ . Since $a > 2\gamma$ , $\gamma \leq \frac{a - 1}{2}$ , and therefore $p^{a - \gamma} - p^{\gamma} \geq p^{\frac{a + 1}{2}} - p^{\frac{a - 1}{2}} = p^{\frac{a - 1}{2}}(p - 1)$ .

Claim: $p^{\frac{a - 1}{2}}(p - 1) > 2a^2$ for odd primes and $a \geq 11$ . If we prove $3^{\frac{a - 1}{2}}\cdot 2 > 2a^2$ , we would be done. Note that for $a = 11$ , $3^5 \cdot 2 = 486 > 242$ , so say we are done for $a = \{11, \dots, k\}$ . Then, $2 \cdot 3^{\frac{k}{2}} = \sqrt{3} \cdot 3^{\frac{a - 1}{2}} \cdot 2 > 2\sqrt{3}k^2 > 2(k + 1)^2$ , since $2\sqrt{3} > 3$ , and $k^2 > 5k > 4k + 1$ . Therefore, we are simply reduced to a case check from $a = 1, 2, \dots 11$ .

Keep in mind that $p \mid a$ and we are only considering $p$ odd at the moment, so for $a = 10, p = 5, 10^{4} + 5^{10} \equiv 2\mod{3}$ , not possible. For $a = 9, p = 3$ , which gives $9^4 + 3^9 = 162^2$ , which works. For $a = 8$ , no odd primes divide it. For $a = 7$ , $p = 7$ and this fails. For $a = 6, p = 3$ and $6^4 + 3^6 = 45^2$ , so this works. For $a = 5, p = 5$ , and this fails. For $a = 4$ , no odd primes divide it. For $a = 3$ , we fail. So, we have the solutions $(9, 3), (6, 3)$ as well.

Now, consider $p = 2$ . We have $2a^2 = 2^{a - \gamma} - 2^{\gamma} \implies a^2 = 2^{\gamma - 1}(2^{a - 2\gamma} - 1)$ . Since $2 \nmid 2^{a - 2\gamma} - 1$ , we have $v_2(a^2) = \gamma - 1$ , which means $2^{\gamma - 1}$ is a perfect square therefore $2^{a - 2\gamma} - 1$ is a perfect square as well. However, this is not possible if $a \geq 2\gamma + 2$ , as then $2^{a - 2\gamma} - 1 \equiv 3\mod{4}$ , so $a = 2\gamma + 1$ , which gives $(2\gamma + 1)^2 = 2^{\gamma - 1}$ , impossible by parity (clearly $\gamma \neq 1$ ).

So, we are done and our solutions are $\boxed{(1, 3), (2, 3), (6, 3), (9, 3)}$ .

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#13 h Oct 12, 2024, 5:15 AM

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Wow okay I hate this problem. We want to solve $p^a + a^4 = n^2$ , or $p^a = (n - a^2)(n + a^2)$ . Let $p^\alpha = n - a^2$ and $p^\beta = n + a^2$ , since $a > 0$ , $\alpha < \beta$ .

We first eliminate the case $p = 2$ . Then, $2^{\alpha - 1}(2^{\beta - \alpha} - 1) = a^2$ , $2^{\beta - \alpha} - 1$ is a perfect square, so $\beta - \alpha = 1$ , by taking modulo $4$ . Therefore, $2(n - a^2) = n + a^2$ , which gives $n = 3a^2$ , or $2^a = 8a^2$ , which clearly has no solutions.

Now suppose $p > 2$ . Then
$p^\beta - p^\alpha = 2a^2 \ge (p - 1)p^{\frac{a - 1}{2}}.$ Now we proceed with a massive cases bash:

$p = 3$ : then $a \le 9$ . If $a = 1$ , $3^1 + 1^4 = 3^2$ works. If $a = 2$ , $3^2 + 2^4 = 5^2$ works. If $a = 3$ , $3^3 + 3^4 = 108$ fails. If $a = 4$ , $3^4 + 4^4 \equiv 2 \pmod 5$ fails. If $a = 5$ , $3^5 + 5^4 = 2^2 \cdot 7 \cdot 31$ fails. If $a = 6$ , $3^6 + 6^4 = 45^2$ works. If $a = 7$ , $3^7 + 7^4 \equiv 12 \pmod{16}$ fails. If $a = 8$ , $3^8 + 8^4 \equiv 3 \pmod 7$ fails. If $a = 9$ , $3^9 + 9^4 = (2 \cdot 3^4)^2$ works.

$p = 5$ : then $a \le 4$ . If $a = 1$ , $5^1 + 1^4 = 6$ fails. If $a = 2$ , $5^2 + 2^4 = 41$ fails. If $a = 3$ , $5^3 + 3^4 = 206$ fails. If $a = 4$ , $5^4 + 4^4 = 881$ fails.

$p \ge 7$ : All $a$ fail.

This post has been edited 1 time. Last edited by KnowingAnt, Oct 12, 2024, 5:17 AM

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#14 h Dec 16, 2024, 3:46 AM

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Write the equation as $p^a+a^4 = k^2$ , or $p^a = (k-a^2)(k+a^2)$ . So both factors are powers of $p$ , i.e.
$2a^2 = p^r - p^{a-r} \geq \begin{cases} p^{a/2 + 1} - p^{a/2 - 1} = p^{a/2 - 1} (p^2-1) & \text{$a$ even} \\ p^{(a+1)/2} - p^{(a-1)/2} = p^{(a-1)/2} (p-1) & \text{$a$ odd} \end{cases}$ Okay, now time for the bash.

Even Case: If $p = 2$ , the inequality holds only for $a \leq 20$ , and no cases yield valid solutions. If $p = 3$ , check $a \leq 6$ , which yields $a = 2$ and $a = 6$ . If $p \geq 5$ , no values of $a$ work.

Odd Case: For $p = 2$ , check $a \leq 23$ ; for $p = 3$ , check $a \leq 12$ , yielding $a = 1$ and $a = 9$ ; for $p = 5$ , check $a \leq 6$ ; for $p \geq 7$ we have $a \leq 4$ which clearly will not yield any solutions.

Thus we have $(1, 3)$ , $(2, 3)$ , $(6, 3)$ , $(9, 3)$ in total.

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#15 h Jan 2, 2025, 9:42 AM

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Junior vibes but time consuming

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N3bula

#16 h Feb 13, 2025, 4:00 AM

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If we have $p^k=n^2-a^4$ we get that $p^k=(n-a^2)(n+a^2)$ thus both $n-a^2$ and $n+a^2$ are powers of $p$ so:
$n-a^2\mid n+a^2$ Thus:
$n-a^2\mid n+a^2+n-a^2=2n$ Thus if $n-a^2\neq 1$ and $n-a^2\neq 2$ we get $p\mid n$ . This means we must also have $p\mid a^2$ , thus $p^2\mid n$ , thus we can divide
by $p^2$ to obtain $p^{k-4}=(\frac{n}{p^2}-(\frac{a}{p})^2)(\frac{n}{p^2}+(\frac{a}{p})^2)$ we can repeat the above argument until we get values $a'$ and $n'$ where $p^i=(n'-a'^2)(n'+a'^2)$ and where we have $n'-a'^2=1$ or $2$ . If $n'-a'^2=1$ we get $p^i=2a'^2+1$ , suppose $p\neq 2$ and $p\neq 3$ , thus we have that $i<a'$ and we also have that $i+4k<a'p^k$ for all $k$ so $p=2$ or $p=3$ , if $p=2$ we have $2^i=2a'^2+1$ which is clearly not possible so $p=3$ is the only possible case. If $a'>2$ we have that $a'>i$ and also that $2a'>4$ so $i+4k<3^ka'$ thus when $a'=1$ we get the solutions $(1, 3)$ and $(9, 3)$ and when $a'=2$ we get the solutions $(2, 3)$ and $(6, 3)$ . If $n'-a'^2=2$ we get $2^i=2a^2+2$ so $2^{i-1}=a^2+1$ this is clearly not possible because of catalans unless $i-1=1$ , if $i-1=1$ we get $a=1$ however this means $a=0$ which is clearly not possible.

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#17 h Mar 11, 2025, 6:25 PM

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uh I hate size

Let $p^a + a^4 = k^2.$ Thus, $p^a = (k - a^2)(k + a^2),$ and assume the factors are $k - a^2 = p^{\ell}, k + a^2 = p^{a - \ell},$ where we require $\ell < a - \ell.$ Thus, we have $a^2 = \frac{p^\ell(p^{a - 2\ell} - 1)}2.$ Clearly, $a > 1.$
Claim: $p = 2, 3$ are the only possible values
Proof: We first handle $p\geq11.$ We have $a^2 = \frac{p^\ell(p^{a - 2\ell} - 1)}2.$ Define the function $f(x) = p^{a - x} - p^x,$ where $x \in [0, \frac a2).$ We have that $f'(x) = -\ln p(p^{a - x} + p^x) < 0.$ Hence, $f$ is decreasing over the interval. Clearly note that $\ell < \frac a2,$ and thus $\ell \leq \frac{a - 1}{2}.$ Thus, for integers $\ell,$ we get $f(\ell) \geq f(\frac{a - 1}{2}) = (p - 1)p^{\frac{a - 1}2} > 2a^2$ for $p\geq11.$

Now, assume $3 < p < 11.$ Observe that from taking modulo $p,$ we get $a^2 \equiv -\frac12\pmod p.$ However,
$\begin{align*} \left(\frac{-2}{p}\right) & = \left(\frac{-1}p\right)\left(\frac{2}{p}\right) \\ & = (-1)^{\frac{p^2-1}{8} + \frac{p - 1}{2}} \\ & = (-1)^{\frac{ p -1}{8}(p + 5)} \end{align*}$ which is a contradiction as the expression evaluates to $-1$ for $p = 5, 7.$

If $p = 2:$ let $a'\in\mathbb Z$ satisfy $a = 2^{\frac{\ell-1}{2}}a'.$ Hence, we require $(a')^2 = 2^{a - 2\ell} - 1.$ However, by catalan, we require $a - 2\ell = 1,$ and thus $a' = 1.$ However, we have that $a$ is a power of 2 while it is an odd number, and so $a = 1.$ This doesn't work by testing it.

If $p = 3:$ We have $2a^2 = 3^{a -\ell} - 3^{\ell} \geq 2\cdot3^{\frac{a - 1}2}$ (recall the lower bound we established earlier). However, $2\cdot3^{\frac{a-1}2} > a^2$ for $a > 11.$ Simple testing tells us that $a = 1, 2, 6, 9$ work.

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#18 h Mar 29, 2025, 11:57 PM

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Anubhab vaiya, we felt that (pun fully intended)

Note $a=1$ gives solution $(1,3)$ . Now we consider $a>1$ .
Case 1: $p \neq 2$
$p^a+a^4=b^2 \implies p^a=(b-a^2)(b+a^2).$ Note $(b-a^2,b+a^2)=(b-a^2,a^2)=$ some power of $p$ since $p \neq 2$ . Now, $\frac{a^2}{b-a^2} \in \mathbb{N} \dots \textbf{(i)}$ . Write $b=p^{n_1}m_1, a=p^{n_2}m_2$ where $(m_1,m_2)=1$ .

$2n_2>n_1,\textbf{(i)} \implies m_1=p^{2n_2-n_1}m_2^2+1$ (since $(p,m_1)=1$ ) $\implies 2a^2+p^{n_1}=p^{a-n_1} \implies 2p^{2n_2-n_1}m_2=p^{a-2n_1}-1$ , contradiction.
$2n_2=n_1, \textbf{(i)} \implies m_1=m_2+1 \implies b-a^2=p^{\bullet} \implies -m_2^2+m_2+1=p^{\bullet -2n_2}$ but for $m_2>1, -m_2^2+m_2+1<0.$ $m_2=1 \implies b+a^2=3p^{2n_2} \implies p=3 \implies 3^{4n_2+1}=3^{3^{n_2}}$ and gives solution $(9,3)$
$2n_2<n_1, \textbf{(i)} \implies p^{n_1-2n_2}m_1=m_2^2+1 \implies 2a^2+p^{2n_2}=p^{a-2n_2} \implies p^{a-4n_2}-1=2m_2^2 \dots \textbf{(ii)}$ . If $n_2=0,1$ we get solution $(2,3), (6,3)$ . If $m_2=1$ , we get solution $(9,3)$ . Now consider $n_2,m_2>1$ . If $m_2<p$ (note the relation cant be that of equality): trivially $3^{n_2}m_2 - 4n_2 > log_p(2)+2 \implies p^{n_2}m_2 - 4n_2 >log_p(2)+2 \implies p^{a-4n_2}>2p^2 \implies p^{a-4n_2}-1 \geq 2p^2 >2m_2^2$ , a contradiction. So, let $m_2=u+p$ where $u>0$ . Trivially $p^{n_2+2} \geq 6n_2p+2p \implies p^{n_2+2}-6n_2p-2p \geq u(2-p^{n_2+1}) \implies \frac{p^{n_2+1}(p+u)-6n_2p-2p}{2u} \geq 1 \implies p^{\frac{p^{n_2+1}(p+u)-6n_2p-2p}{2u}} \geq 3>e>(1+ \frac{u}{p})^{\frac{p}{u}} \implies p^{p^{n_2+1}m_2-6n_2p} > m^{2p} \implies p^{\frac{a-6n_2}{a-4n_2}} > m^{\frac{2}{a-4n_2}} \implies 1 > p^{\frac{6n_2-a}{a-4n_2}}m_2^{\frac{2}{a-4n_2}}$ (trivially $6n_2-a<0$ for $n_2>1$ ) $\implies p^{a-4n_2} > p^{2n_2}m_2^2>2m_2^2 \implies p^{a-4n_2}-1>2m_2^2$ , a contradiction from $\textbf{(ii)}$ .
Case 2: $p=2$
Subcase 1: $2 \nmid a$ . Let $b-a^2=2^{t_1} \implies 2^{t_1}+2a^2=2^{t_2} \implies t_1=1$ (since $t_1 \neq t_2$ ) $\implies (b+a^2)=2(a^2+1)=2^{\bullet}$ , contradiction by zsigmondy
Subcase 2: $2 \mid a$ . Let $a=2^tk$ where $(2,k)=1$ . So, $2^{2^tk-4t}+k^4=(\frac{b}{2^{2t}})^2$ . When $2^tk-4t>0$ this becomes subcase 1. Trivially, $k \geq 3 \implies 2^tk-4t > 0$ . If $k=1$ then trivially, except for $t \in \{1,2,3,4\}$ , $2^tk-4t>0$ . Case working thru $k=1, t \in \{1,2,3,4\}$ gives no solution.

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