Prove that there no solutions for

\[ \large{5^a+8^b=c^2 }\]

For \(a,b,c \in \mathbb{Z^{+}} \) and \(b\) is not an even positive integer.

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## Comments

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TopNewest\[c^2 \equiv 0, 1, 4, 9, 6, 5 \pmod{10}\]

\[5^a \equiv 5 \pmod{10}\]

\[8^b \equiv 8, 2 \pmod{10} (b \text{ is odd})\]

\[5^a + 8^b \equiv 3,7 \pmod{10}\]

Clearly, the first statement and the last statement share no residues. Thus proven.

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Haha, the same way I solved the question!

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Nice problem. By the way , from where have you collected these problems?

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What's mod? And how are 6 and 5 also included?

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\[ a \equiv b \pmod c\]

Implies a leaves same remainder as b when divided by c.

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yes it must have i;e solve by indicies a+b=2

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The bases are not same. Additionally, there's no multiplication involved so you cannot add the indices.

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