Proof Contest Day 1

Prove that there no solutions for

5a+8b=c2 \large{5^a+8^b=c^2 }

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

Note by Department 8
3 years, 8 months ago

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c20,1,4,9,6,5(mod10)c^2 \equiv 0, 1, 4, 9, 6, 5 \pmod{10}

5a5(mod10)5^a \equiv 5 \pmod{10}

8b8,2(mod10)(b is odd)8^b \equiv 8, 2 \pmod{10} (b \text{ is odd})

5a+8b3,7(mod10)5^a + 8^b \equiv 3,7 \pmod{10}

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

Sharky Kesa - 3 years, 8 months ago

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Haha, the same way I solved the question!

Department 8 - 3 years, 8 months ago

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

Priyanshu Mishra - 3 years, 8 months ago

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@Priyanshu Mishra Own thinking. Try other problems too from this set.

Department 8 - 3 years, 8 months ago

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@Department 8 Trying others.

Priyanshu Mishra - 3 years, 8 months ago

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

Yuki Kuriyama - 3 years, 8 months ago

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ab(modc) a \equiv b \pmod c

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

Department 8 - 3 years, 8 months ago

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

amar nath - 3 years, 8 months ago

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

Arulx Z - 3 years, 7 months ago

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