It is easy to show that both $m$ and $n$ are even.Let $m=2r$ and $n=2s$

Hence, $2^{2r}=a^{2}-3^{2s}=(a-3^{s})(a+3^{s})$

Hence, $a-3^{s}=2^{i}$...$(1)$ and $a+3^{s}=2^{2r-i}$...$(2)$

$(2)-(1)$ gives $2.3^{s}=2^{i}(2^{2r-2i}-1)$, which implies $i=1$.

Thus, $a-3^{s}=2$ and $a+3^{s}=2^{2r-1}$.Hence, $3^{s}=2^{2r-2}-1$...$(3)$

Suppose, $s>1$.Then $r≥3$.But then the equation $(3)$ is impossible since when divided by $8$, the left hand side $3^{s}$ leaves a remainder $1$ or $3$ while the right hand side would leave the remainder $7$.Thus $s=1$ is the only possibility.When $s=1$,i.e, $n=2$,we have the solution $2^{4}+3^{2}=25$.Thus $(m,n)=(4,2)$ is the only solution.

$a^{2}≡0 or 1(mod 3)$ and $2^{m}+3^{n}≡2^{m}(mod 3)$.But $2^{m}$ is not congruent $0$ modulo $3$.

So,$2^{m}≡1(mod 3)$ which implies $m$ is even.Hence, $3^{n}≡a^{2}≡0 or 1 (mod 4)$.But $4$ does not divides $3^{n}$.So,$3^{n}≡1(mod4)$ which implies $n$ is even.

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## Comments

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TopNewestSuppose $2^{m}+3^{n}=a^{2}$.

It is easy to show that both $m$ and $n$ are even.Let $m=2r$ and $n=2s$

Hence, $2^{2r}=a^{2}-3^{2s}=(a-3^{s})(a+3^{s})$

Hence, $a-3^{s}=2^{i}$...$(1)$ and $a+3^{s}=2^{2r-i}$...$(2)$

$(2)-(1)$ gives $2.3^{s}=2^{i}(2^{2r-2i}-1)$, which implies $i=1$.

Thus, $a-3^{s}=2$ and $a+3^{s}=2^{2r-1}$.Hence, $3^{s}=2^{2r-2}-1$...$(3)$

Suppose, $s>1$.Then $r≥3$.But then the equation $(3)$ is impossible since when divided by $8$, the left hand side $3^{s}$ leaves a remainder $1$ or $3$ while the right hand side would leave the remainder $7$.Thus $s=1$ is the only possibility.When $s=1$,i.e, $n=2$,we have the solution $2^{4}+3^{2}=25$.Thus $(m,n)=(4,2)$ is the only solution.

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Thank you for this solution! Been thinking about this all day. Never got the chance to sit down with a pen and paper, unfortunately...

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How to show that both $m$ and $n$ are even?

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$a^{2}≡0 or 1(mod 3)$ and $2^{m}+3^{n}≡2^{m}(mod 3)$.But $2^{m}$ is not congruent $0$ modulo $3$.

So,$2^{m}≡1(mod 3)$ which implies $m$ is even.Hence, $3^{n}≡a^{2}≡0 or 1 (mod 4)$.But $4$ does not divides $3^{n}$.So,$3^{n}≡1(mod4)$ which implies $n$ is even.

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How about $(3,0)$ and $(0,1)$ ?

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m, n are positive integers. 0 is not positive.

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Could you please explain your solution from line 5 onwards? How is it 2.3^s? Thanks.

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I just subtracted eqn $1$ from eqn $2$ and got $2\times3^{s}=2^{2r-i}-2^{i}=2^{i}(2^{2r-2i}-1)$.

Hence, $3^{s}=2^{i-1}(2^{2r-2i}-1)$.If $i>1$,then $2$ divides $3^{s}$, which is impossible.So $i=1$.

So,$3^{s}=2^{1-1}(2^{2r-2.1}-1)=2^{2r-2}-1$, which is eqn $3$.Then the solution is very clear.

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(4,2),(3,0),(0,1),

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