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Is this valid for all cases? $${\underbrace{aaa\dots a}_\text{$2n$ times}}^{\overbrace{bbb\dots b}^\text{$2n$ times}}+{\underbrace{bbb\dots b}_\text{$2n$ times}}^{\overbrace{aaa\dots a}^\text{$2n$ times}} \equiv 0 \mod (a+b)$$

For example, this holds in case of $2222^{5555}+5555^{2222}$, which is divisible by $7$, however I'm getting confused on how to prove it through congruence.

Any ideas or hints?

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    $\begingroup$ You tagged "contest-math". Hope this is not an ongoing contest ! $\endgroup$ – Peter May 8 at 8:54
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    $\begingroup$ Oh not at all. It's actually a practice question in my book. Sorry for the ambiguity. $\endgroup$ – user907745 May 8 at 9:03
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    $\begingroup$ $$22^{33}+33^{22}\equiv 2^1+3^2\equiv2+4\equiv 1\pmod 5$$ $\endgroup$ – miracle173 May 8 at 9:12
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    $\begingroup$ I rolled back you question to your initial question Please do not change your question because this invalidates my answer. If you have another question then pose a new one. $\endgroup$ – miracle173 May 8 at 19:03
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    $\begingroup$ @Crease You say that this is a practice problem from a book. Which book? $\endgroup$ – Xander Henderson May 8 at 19:41
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We have \begin{align} 2222^{5555}+5555^{2222}&\equiv3^{5\times1111}+4^{2\times1111}&&\pmod7\\ &\equiv243^{1111}+16^{1111}&&\pmod7\\ &\equiv(-2)^{1111}+2^{1111}&&\pmod7\\ &\equiv0&&\pmod7 \end{align} You can also see this for similar such thing.

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That is wrong!

$$22^{33}+33^{22}\equiv 2^1+3^2\equiv2+4\equiv 1\pmod 5$$

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    $\begingroup$ apologies, I don't quite understand this answer; I'm sure I'm missing something, but how does it address the question? $\endgroup$ – Atticus Stonestrom May 8 at 14:02
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    $\begingroup$ @AtticusStonestrom the OP changed the question after I answered it. In the meantime I rolled it back $\endgroup$ – miracle173 May 8 at 19:06
  • $\begingroup$ ah, thank you, that makes much more sense! (+1) $\endgroup$ – Atticus Stonestrom May 8 at 21:45
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By Fermat's little theorem, we reduce the exponents $\pmod6$.

We use CRT via Bezout, and we have $5555\equiv1\pmod2$ and $5555\equiv2\pmod3$. So $5555\equiv5\pmod6$, since $(-1)2+(1)3=1\implies 5555\equiv -2\cdot2+3\cdot1\equiv5\pmod6$.

Next $2222\equiv0\pmod2$ and $2222\equiv2\pmod3$ so $2222\equiv2\pmod6$.

Thus $5555^{2222}+2222^{5555}\equiv4^2+3^5\equiv 16+243\equiv259\equiv0\pmod7$, using what you got.

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