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# Proof Contest Day 1

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.

Note by Lakshya Sinha
1 year, 6 months ago

## Comments

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

$5^a \equiv 5 \pmod{10}$

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

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

Clearly, the first statement and the last statement share no residues. Thus proven. · 1 year, 6 months ago

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Haha, the same way I solved the question! · 1 year, 6 months ago

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Nice problem. By the way , from where have you collected these problems? · 1 year, 6 months ago

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Own thinking. Try other problems too from this set. · 1 year, 6 months ago

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Trying others. · 1 year, 6 months ago

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What's mod? And how are 6 and 5 also included? · 1 year, 6 months ago

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$a \equiv b \pmod c$

Implies a leaves same remainder as b when divided by c. · 1 year, 6 months ago

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yes it must have i;e solve by indicies a+b=2 · 1 year, 6 months ago

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The bases are not same. Additionally, there's no multiplication involved so you cannot add the indices. · 1 year, 5 months ago

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