10
$\begingroup$

Solve $$7^x+x^4+47=y^2$$ where $x, y \in \mathbb{N}^*$

If $x$ is odd then the left term is congruent with $3$ mod $4$ so it couldn't be a perfect square, so we deduce that $x=2a$ and the relation becomes $$49^a+16a^4+47=y^2$$ and it is easy to see that the left term is divisible by $16$ so we obtain that $y=4b$, so we have to find $a$ and $b$ such that $$49^a+16a^4+47=16b^2$$From this point I was completely stuck. I think that there are no solutions but how can I prove it?

$\endgroup$
2
  • $\begingroup$ The next square after $(7^a)^2$ is probably more than the LHS $\endgroup$
    – Empy2
    Commented Sep 26, 2018 at 12:36
  • $\begingroup$ @Empy2 Yes you are right Sir $\endgroup$
    – razvanelda
    Commented Sep 26, 2018 at 12:43

3 Answers 3

Reset to default
6
$\begingroup$

If $x$ is even and let $x=2a$, then $(7^a)^2<7^x+x^4+47<(7^a+1)^2=7^x+2\times7^a+1$ if $(2a)^4+47<2\times7^a+1$, which is true for $a \ge 4$. Therefore, it is enough to consider only $x=2, 4$ and $6$.

$\endgroup$
1
  • $\begingroup$ Yeap, thank you $\endgroup$
    – razvanelda
    Commented Sep 26, 2018 at 12:39
2
$\begingroup$

You can't prove no solutions. $x=4, y=52$ is a solution.

$\endgroup$
3
  • 1
    $\begingroup$ Ok, so my thoughts were wrong. Still, how do we find all solutions? $\endgroup$
    – razvanelda
    Commented Sep 26, 2018 at 12:28
  • $\begingroup$ To be honest, I don't know how to find all solutions. A computer check (using Python) gave no other solutions up to $x=1000$. $\endgroup$
    – paw88789
    Commented Sep 26, 2018 at 12:33
  • $\begingroup$ you have to check only a finte number of $x$. I think $100$ is enough $\endgroup$
    – Exodd
    Commented Sep 26, 2018 at 12:33
0
$\begingroup$

When $x$ is odd then $7^x+x^4+47\equiv 3(\mod 4)$. So, it is not possible $y^2\equiv 3(\mod 4)$.

Let $x=2k$, then for $k\geq 4$

$$(7^k)^2<7^{2k}+(2k)^4+47<(7^k+1)^2$$

This mean we have $k\leq 3$. if try to $k=1,2,3$ only $x=4$ is a solution.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .