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#1 h Apr 16, 2025, 4:42 PM

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Find all pairs of positive integers $(m,n)$ satisfying $2^m+21^n$ is a perfect square

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#3 h Apr 16, 2025, 6:37 PM

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The only solution is $m = 2$ and $n = 1$ . This clearly works. Now we prove it's the only solution. By mod $3$ , $m$ has to be even, so let $m = 2k$ . Additionally, let the expression be $x^2$ . We have $21^n = (x - 2^k)(x + 2^k)$
Case 1: $x - 2^k = 1$ .
Then $x + 2^k = 21^n$ , so $2^{k + 1} + 1 = 21^n$ , which is impossible by Mihailescu/Zsigmondy.
Case 2: $7 \mid x - 2^k$ .
Then since $7$ doesn't divide $(x + 2^k) - (x - 2^k)$ , $x + 2^k$ has to be a power of $3$ . Since $x + 2^k > 1$ , $3 \mid x + 2^k$ , so $3 \nmid x - 2^k$ . Thus, $x - 2^k$ is a power of $7$ . Since $x - 2^k$ and $x + 2^k$ multiply to $21^n$ , we have $x - 2^k = 7^n$ and $x + 2^k = 3^n$ , which fails due to size.

Case 3: $3 \mid x - 2^k$ .
Then since $3$ doesn't divide $(x + 2^k) - (x - 2^k)$ , $x + 2^k$ has to be a power of $7$ . Since $x + 2^k > 1$ , $7 \mid x + 2^k$ , so $7 \nmid x - 2^k$ . Thus, $x - 2^k$ is a power of $3$ . Since $x - 2^k$ and $x + 2^k$ multiply to $21^n$ , we have $x - 2^k = 3^n$ and $x + 2^k = 7^n$ . This implies that $7^n - 3^n$ is a power of $2$ . By Zsigmondy's Theorem, for any $n > 1$ , we can choose a prime $p$ dividing $7^n - 3^n$ so that $p$ does not divide $7^1 - 3^1 = 4$ , and therefore $7^n - 3^n$ is not a power of $2$ . Hence $n = 1$ , so $2^{k + 1} = 4 \implies k = 1$ , giving $m = 2$ .

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KAME06

#4 h Apr 16, 2025, 6:49 PM

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We have that $2^m+21^n=x^2$ . Working mod 3: $2^m+21^n \equiv (-1)^m =x^2 \pmod{3}$ . $-1$ isn't a quadratic residue, so $m$ must be even.
If $m=2b$ , then $2^{2b}+21^n=(2^b)^2+21^n=x^2 \Rightarrow 21^n=(x-2^b)(x+2^b)$ .
If there is a prime $p \neq 2$ such $p \mid x-2^b$ and $p \mid x+2^b$ , but that would mean $p \mid x+2^b-(x-2^b)=2^{b+1}$ . Contradiction.
With that, and knowing that $x+2^b>x-2^b$ and $21^n=3^n7^n$ , we have that:
$\begin{cases} x+2^b=7^n \\ x-2^b=3^n \end{cases}$ $\Rightarrow x+2^b-(x-2^b)=2^{b+1}=7^n-3^n$
Working mod 5: $2^{b+1} = 7^n-3^n \equiv 7^n-(-7)^n \pmod{5}$ , so $n$ must be odd because $5 \nmid 2^{b+1}$ .
Suppose that $b>1$ , then $8=2^3 \mid 2^{b+1} \Rightarrow 2^{b+1}=7^n-3^n \equiv (-1)^n-3^n \equiv -1-3 =-4 \equiv 0 \pmod{8}$ (that last equivalence because $n$ is odd). Contradiction. Then $b=1 \Rightarrow 4=7^n-3^n$ . Knowing that $7^n$ grows faster than $3^n$ , the unique answer is $n=1$ . So we have $(m, n)=(2, 1)$
$2^2+21^1=4+21=25=5^2$ .
Edit: Forgot one case: If $x-2^b=1$ , then $21^n=2^{b+1}+1 \Rightarrow 7 \mid 2^{b+1}+1$ , but the residues mod 7 of $2^{b+1}$ are $2, 4, 1$ . Contradiction.

This post has been edited 4 times. Last edited by KAME06, Apr 16, 2025, 6:57 PM

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#5 h Apr 16, 2025, 7:14 PM

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There might be some overlap with the previous posts but I'll post my solution anyway.
Mod 3 we have $(-1)^m \equiv \Box$ hence $m$ is even.
If $m=2$ then $21^n+4=k^2\implies 21^n=(k+2)(k-2)$ . Then $k-2,k+2$ are coprime and $k>3$ hence $k+2=7^n,k-2=3^n\implies 3^n+4=7^n$ . If $3^n+4\le 7^n$ then $3^{n+1}+4<3(3^n+4)\le 3\cdot 7^n<7^{n+1}$ hence $n=1$ is the only solution. This gives $21+4=25$ , from now on $m\ge 4$ .

Then mod 8 we have $(-3)^n \equiv 2^m+21^n\equiv \Box$ hence $n$ is even and $n=2\nu$ .
Then $2^m+21^{2\nu}=k^2\implies 2^m=(k+21^\nu)(k-21^\nu)$ . Now $(k+21^\nu,k-21^\nu)=(k+21^\nu,k-21^\nu,2\cdot 21^\nu)=2$ hence $k-21^\nu=2, k+21^\nu=2^{m-1} \implies 21^\nu=2^{m-2}-1$ .
Then $m>4$ hence $m\ge 6$ and we have $(-3)^\nu\equiv -1(8)$ . The only powers of $-3\mod 8$ are $-3,1$ , contradiction.

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