2
$\begingroup$

I noticed a strange relation months ago :

$\begin{cases}3^5+10^2=7^3\\3+7=10\\2+3=5\end{cases}$

For the sake of math, I searched for positive integer non trivial (i.e. not containing any 0) solutions of this system:

$\begin{cases}x^a+y^b=z^c\\x+z=y\\b+c=a\end{cases}$

So far I've dug with Python in brute-force to $x,y,z\in [[1,100]]\ \mbox{and}\ a,b,c\in [[1,30]]$ and the only tuples $(x,y,z,a,b,c)$ I've found are $(1, 3, 2, 3, 1, 2)$, $(3, 9, 6, 3, 1, 2)$ and $(3, 10, 7, 5, 2, 3)$.

I consider really strange this relation, even weirder since I can't find out more without devoting time to it. Could there be more solutions?

I think not: given a triplet $(a,b,c)$ such that $b+c=a$, only few integers $(x,y,z)$ verify both conditions.

EDIT (update) : Still no other solution for $x,z\in [[1,2000]]\ \mbox{and}\ b,c\in [[1,50]]$.

Improve this question
$\endgroup$
5
  • 1
    $\begingroup$ I voted for this question to be closed. I think it falls under mathoverflow.net/help/dont-ask because of the way it is phrased. However, a number of slight tweaks such as "Are there infinitely many such solutions?" might make this question more appropriate for this site. $\endgroup$
    – Neil Hoffman
    Feb 21, 2017 at 21:07
  • $\begingroup$ OK, I edited my question, you may remove your vote. $\endgroup$
    – Maxence Seymat
    Feb 21, 2017 at 21:19
  • 1
    $\begingroup$ Not a complete solution, but a start. First, you might as well eliminate two of the variables, say $a$ and $y$, from the three equations. Your problem then reduces to finding positive integer solutions to single equation $$x^{b+c}+(x+z)^b=z^c.$$ This equation yields $$ x^{b+c}+x^b\equiv0\pmod{z}\quad\text{and}\quad z^b\equiv z^c\pmod{x}. $$ Let's look for solutions with $\gcd(x,z)=1$. Then the congruences give $$ x^c\equiv-1\pmod{z}\quad\text{and}\quad z^{|b-c|}\equiv 1\pmod{x}.$$ $\endgroup$
    – Joe Silverman
    Feb 22, 2017 at 2:12
  • $\begingroup$ I also analysed a bit the equation, and the solutions to the reduced single equation must verify $x<z$ and $c>b$ since $z^c=x^{b+c}+(x+z)^b>x^c$ and $z^c=x^{b+c}+(x+z)^b>z^b$. $\endgroup$
    – Maxence Seymat
    Feb 22, 2017 at 16:29
  • $\begingroup$ I found interesting rewriting $z=x+k$ and $c=b+i$ with $k,i\in\mathbb{N}^∗$. With your work, it becomes $x^{2b}(x^i+1)\equiv0\ [z]\ \mbox{and}\ k^b(k^i-1)\equiv0\ [x]$. Too bad I didn't think to congruences earlier. $\endgroup$
    – Maxence Seymat
    Feb 22, 2017 at 18:52

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.