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I'm trying to solve this problem:

Suppose $a$ and $b$ are positive integers such that $a^2 + b^2$ is even and $a^3 + b^3$ is a multiple of 3. What is the largest positive integer that must divide $a^5 + b^5$?

My work so far is as follows:

If $a^2 + b^2$ is even, then $a$ being even implies $b$ being even, and $a$ being odd implies $b$ being odd (and vice versa). That is, $a \equiv b$ (mod 2).

If $a^3 + b^3$ is a multiple of 3: First, by Fermat's Little Theorem, we have $a^3 \equiv a$ (mod 3) and vice versa for $b$. As a result, we can conclude that $a + b \equiv 0$ (mod 3).

However, I don't know how to handle the $a^5 + b^5$. I've tried multiplying $(a^3 + b^3)(a^2 + b^2) = (a^5 + b^5) + a^2b^2(a + b)$, but that didn't seem to help. Can someone give me a few pointers on where to proceed?

Note: Testing a few examples, I think the answer is 6 ($(a, b) = (3, 3)$ yields $486$, $(a, b) = (2, 4)$ yields $1056$, ...). Rather than just the answer to this problem, I'm more concerned about the proof methods that I should develop to tackle problems similar to this in the future. I'd greatly appreciate if someone could provide a proof, relying mainly on theorems/lemmas related to modular arithmetic, that leads to the answer to this problem.

Bill Dubuque
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Christopher Miller
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    Examples are always good. At least you can discover the answer empirically. That makes it easier to solve. (Note: in this case, it is easy to go from the empirical result to a full proof). – lulu Aug 04 '23 at 20:00
  • @lulu Is my approach of multiplying $(a^3 + b^3)(a^2 + b^2)$ in the correct direction? I've been stuck on this for a while... – Christopher Miller Aug 04 '23 at 20:08
  • Again, start with examples. – lulu Aug 04 '23 at 20:19
  • Just compute the $\gcd$ of the values of $a^5+b^5$ for a couple of those solutions. The largest divisor that is sure to divide all must divide the $\gcd$ of any two. For example $\gcd(3^5+3^5, 4^5+2^4)$. That would prove that the number you want divides $6$. – NDB Aug 04 '23 at 20:32
  • For the other direction, solve the system $\begin{cases}a^2+b^2&\equiv0\pmod{6}\a^3+b^3&\equiv0\pmod{6}\end{cases}$ and verify that for those $a^5+b^5\equiv0\pmod{6}$. That proves that $6$ divides all. Note that your examples are almost all solutions of this system already. The other is $(5,1)$ and swapping the values because the system is symmetric. – NDB Aug 04 '23 at 20:35
  • $2,3\mid a+b\iff 6\mid a+b$ by CCRT. So $\bmod 6!:\ a\equiv -b\Rightarrow a^5\equiv (-b)^5\equiv -b^5,$ by the Congruence Power Rule. – Bill Dubuque Aug 04 '23 at 20:51
  • @BillDubuque Sorry, I don't see how this proves the answer? If $6 | a + b$ we can show that $6 | a^5 + b^5$ using the equation I tried working on in the question, but I don't see how you prove that $6$ is the maximum number to divide $a^5 + b^5$. – Christopher Miller Aug 04 '23 at 20:53
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    To show its the max look at the first few values and compute their gcd, as suggested above (a standard technique, e.g. see "Generally" here or see here). – Bill Dubuque Aug 04 '23 at 20:54
  • i.e. $,d\mid 486,1056\Rightarrow d\mid \gcd(486,1056) = 6\ \ $ – Bill Dubuque Aug 04 '23 at 21:01

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