What I did so far:
$333^{555}+555^{333} = [(111×3)^{111}]^5+[(111×5)^{111}]^3$
$[(111×3)^{111}]^5+[(111×5)^{111}]^3 = 111^{555}×3^{555}+111^{333}×5^{333}$
I'm stuck.
Prove that $333^{555}+555^{333}$ is a multiple of $97$
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algebra-precalculus
elementary-number-theory
proof-verification
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6It's not true, double check the statement – 2017-01-09
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0111 is not a multiple of 97. 3^n is also not a multiple of 97 – 2017-01-09
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0It is a multiple of 48. – 2017-01-09
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0I get a remainder of $33$ when that number is divided by $97$. – 2017-01-09
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0Probably $97$ should be $37$. Then it's trivially true by $\,37\mid 111\mid 333,555\ \ $ – 2017-01-09
1 Answers
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It is not true that it's a multiple of $97$ since:
$$\begin{cases}333 \equiv 42 \pmod{97} \\ 555 \equiv 70 \pmod{97} \\ 333 \equiv 45 \pmod{\varphi({97})} \\ 555 \equiv 75 \pmod{\varphi({97})}\end{cases}$$
$$\begin{cases}42^{75} \equiv 45 \pmod{97} \\ 70^{45} \equiv 85 \pmod{97}\end{cases}$$
$$45+85 \not \equiv 0 \pmod{97}$$
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2Better to let the OP correct the problem before posting a counterexample to a typo. – 2017-01-09
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1This is the actual problem on the book, no typos. Thank you for clarifying that. – 2017-01-09