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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