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I came across the following result earlier: $$(4^7 + 2^7) = 0 \pmod 6$$ Which I've seen occur in other forms before, leading me to speculate that there exist numbers n and x such that $$a^x + b^x \equiv a + b \pmod{n}$$ However, I do not know how to prove this, let alone figure out which numbers satisfy this property. Thoughts?

Arturo Magidin
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Alexandra
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  • If $n$ is prime and neither $a$ nor $b$ are divisible by $n$, then $a^{n} + b^{n} \equiv a + b \bmod n$. Then the congruence holds for all $a$ and $b$ (by FLT). If $n$ can be factored into $n = p_1^{k_1} \cdots p_m^{k_m}$ where each $p_i$ is prime, then it suffices to consider the congruence modulo each of these prime powers. The given congruence holds if it holds for each $p_i^{k_i}$ by the CRT. I suspect you could break it into more cases depending on $n$. There may even be a general answer, I do not know. – codeing_monkey Apr 10 '24 at 18:16
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    @codeing_monkey If $n$ is prime then the result holds for all $a$ and $b$ with $x=n$, by Fermat's Little Theorem. – Arturo Magidin Apr 10 '24 at 18:19
  • @ArturoMagidin Oops, thanks. – codeing_monkey Apr 10 '24 at 18:21
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    Euler's theorem says that for $a,n$ coprime, $a^{\varphi(n)} \equiv 1 \pmod{n}$, where $\varphi(n)$ is Euler's totient function, the number of numbers less than $n$ that are coprime to $n$. Using this, for any $a,b$ coprime to $n$, your equation holds for $x=\varphi(n)+1$ – Quick_Fix Apr 10 '24 at 18:35
  • @CameronBuie What is true, as has been mentioned, is that if $x\equiv 1\pmod{\varphi(n)}$ and $\gcd(n,ab)=1$, then it holds. – Arturo Magidin Apr 11 '24 at 00:40
  • @ArturoMagidin Thank you for setting me straight. I should know better than to trust my symbolic memory alone, at this point. – Cameron Buie Apr 11 '24 at 01:38
  • $k^7\equiv k\pmod{6}$ for all integers $k,,$ by $n=6$ in the linked dupe. $\ \ $ – Bill Dubuque Apr 13 '24 at 02:36

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This is a comment not an answer. I saw the following problem in the book Number theory by Sierpinski:

Find smallest composite solutions for $n$ such that $n|2^n-2$ and $n|3^n-3$.

The solution is $n=561=3\times 11\times 17$ so that $ 2^{561}\equiv 2\bmod 561$ and $3^{561}\equiv 3 \bmod{561}$ which results in

$$2^{561}+3^{561}\equiv 2+3\bmod 561$$

Another solution is $n=1105=5\times 13\times 17$ so $$2^{1105}+3^{1105}\equiv 2+3\bmod 1105$$

For $a=2$ and $b=3$, there may be many $n$ to satisfy $a^n+b^n\equiv a+b\bmod n$. It can be shown that for every natural number $a$ there are infinitely many even numbers $n$ such that $n|a^n-a$. Suppose we also have numbers $b$ and $m$ such that $m|b^m-b$ and also $(m, n) = p_1p_2 \cdots$, then we have:

$$a^{p_1p_2\cdots}+b^{p_1p_2\cdots}\equiv a+b\bmod p_1 p_2\cdots$$

Update: From the same book:

It can be shown that for any natural number $a>1$, there are infinitely many natural numbers $n$ such that $n|a^n+1$.

Suppose for another natural number $b$ we also have $n|b^n+1$, then we may write:

$$a^n\equiv -1 \bmod n\Rightarrow a^{n+1}\equiv -a \bmod n$$

$$b^n\equiv -1 \bmod n\Rightarrow b^{n+1}\equiv -b \bmod n$$

Which gives:

$$a^{n+1}+b^{n+1}\equiv-(a+b)\bmod n$$

If this is true then we can conclude that:

$$a^{n+1}+b^{n+1}\equiv|(a+b)|\bmod n$$

Here $x=n+1$.

sirous
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