Today, I came up with the following problem when trying to solve this.

Are there distinct integers $a,b,x,y>1$ such that the equation $$a^b+b^a=x^y+y^x$$ holds? That is,

  • Is there ever an integer that can be written as $x^y+y^x$ in more than one way?

I claim that the answer to this question is 'no'.

I think solving this is beyond my knowledge, though we could start by observing the last digit of each term. The simplest case is to consider the powers of $1,5,6,0$, since they end in those same digits. For example, $$\begin{cases}a\equiv5\pmod{10}\\b\equiv6\pmod{10}\end{cases}\implies a^b+b^a\equiv1\pmod{10}$$ However, this brings an issue, since there is hardly any indication as to what values $x$ and $y$ can take other than them having opposite parity.

PARI/GP code is

  intfun(a,b,x,y)={for(i=2,a,for(j=2,b,for(k=2,x,for(l=2,y,if(i<>k && i<>l && j<>k && j<>l && i^j+j^i-k^l-l^k==0,print(i," ",j," ",k," ",l))))));}

No solutions have been found up to $a,b,x,y\le100$.

  • 1
    $\begingroup$ We can accelerate the search by assuming $a\le b, x\le y, a\le x$ $\endgroup$ – Peter Jul 7 '19 at 19:44
  • $\begingroup$ Still no solution in the range $a,b,x,y\le 200$ $\endgroup$ – Peter Jul 7 '19 at 20:03
  • $\begingroup$ Fermat's little theorem will help. $\endgroup$ – Roddy MacPhee Jul 7 '19 at 23:41
  • $\begingroup$ $a^b-x^y=b^a-y^x$ etc mean they either pump out factors on both sides, or are all coprime. $\endgroup$ – Roddy MacPhee Jul 8 '19 at 1:31
  • $\begingroup$ $x^y-a^b$ rather. $\endgroup$ – Roddy MacPhee Jul 8 '19 at 1:42

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