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.

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. – Sharky Kesa · 8 months, 3 weeks ago

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– Lakshya Sinha · 8 months, 3 weeks ago

Haha, the same way I solved the question!Log in to reply

– Priyanshu Mishra · 8 months, 3 weeks ago

Nice problem. By the way , from where have you collected these problems?Log in to reply

– Lakshya Sinha · 8 months, 3 weeks ago

Own thinking. Try other problems too from this set.Log in to reply

– Priyanshu Mishra · 8 months, 3 weeks ago

Trying others.Log in to reply

– Yuki Kuriyama · 8 months, 2 weeks ago

What's mod? And how are 6 and 5 also included?Log in to reply

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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– Arulx Z · 7 months, 3 weeks ago

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