Prove that there are infinitely many distinct pairs (a, b) of co-prime integers $a > 1$ $a>1$ and $b > 1$ $b>1$ such that $a^{b} + b^{a}$ $a^b+b^a$ is divisible by $a + b$ $a+b$ ?

Question

1790 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-20T12:44:15

See below.

Making $a = 2 k + 1$ $a=2k+1$ and $b = 2 k + 3$ $b=2k+3$ we have that

$a^b+b^a equiv 0 mod (a+b)$ and for $k in NN^+$ we have that $a$ and $b$ are co-primes.

Making $k+1=n$ we have

$(2n-1)^(2n+1)+(2n+1)^(2n-1) equiv 0 mod 4$ as can be easily shown.

Also can be easily shown that
$(2n-1)^(2n+1)+(2n+1)^(2n-1) equiv 0 mod n$ so

$(2n-1)^(2n+1)+(2n+1)^(2n-1) equiv 0 mod 4n$ and thus is demonstrated that for $a=2k+1$ and $b=2k+3$

$a^b+b^a equiv 0 mod (a+b)$ with $a$ and $b$ co-primes.

The conclusion is

... that there are infinitely many distinct pairs $(a, b)$ of co-prime integers $a>1$ and $b>1$ such that $a^b+b^a$ is divisible by $a+b$ .

Prove that there are infinitely many distinct pairs (a, b) of co-prime integers a>1a>1a>1 and b>1b>1b>1 such that a^b+b^aab+baa^b+b^a is divisible by a+ba+ba+b?