# Solve $7^x+x^4+47=y^2$

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?

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

## 3 Answers

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 $$6$$.

• Yeap, thank you Commented Sep 26, 2018 at 12:39

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

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

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.