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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
8 months, 3 weeks ago

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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. Sharky Kesa · 8 months, 3 weeks ago

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@Sharky Kesa Haha, the same way I solved the question! Lakshya Sinha · 8 months, 3 weeks ago

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@Lakshya Sinha Nice problem. By the way , from where have you collected these problems? Priyanshu Mishra · 8 months, 3 weeks ago

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@Priyanshu Mishra Own thinking. Try other problems too from this set. Lakshya Sinha · 8 months, 3 weeks ago

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@Lakshya Sinha Trying others. Priyanshu Mishra · 8 months, 3 weeks ago

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@Sharky Kesa What's mod? And how are 6 and 5 also included? Yuki Kuriyama · 8 months, 2 weeks ago

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@Yuki Kuriyama \[ a \equiv b \pmod c\]

Implies a leaves same remainder as b when divided by c. Lakshya Sinha · 8 months, 2 weeks ago

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yes it must have i;e solve by indicies a+b=2 Amar Nath · 8 months, 2 weeks ago

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@Amar Nath The bases are not same. Additionally, there's no multiplication involved so you cannot add the indices. Arulx Z · 7 months, 3 weeks ago

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