# Are there any nonzero rational solutions to this equation?

Are there any nonzero rational solutions to the equation

$$y^2 = 64x^n + 1$$ where $n\geq 3$ is an integer ?

For the case $n=3$, the question can be settled by basic ideas of elliptic curves, but i'm not sure of the higher cases ?

• $n=4$ is also elliptic curve. – joro Oct 2 '16 at 15:32
• According to sage, there are no nonzero solutions for n=4 and only finitely for n=3. – joro Oct 2 '16 at 15:35
• A result of Fermat (squares are not the sum of two biquadrates) gives that there are no solutions for n=4k. Gerhard "Don't Know About Bi Hexrates" Paseman, 2016.10.02. – Gerhard Paseman Oct 2 '16 at 15:44
• I claim that there are no such solutions for $n\geq4$. – T. Amdeberhan Oct 2 '16 at 16:43
• What is the motivation for considering this particular equation? – RP_ Oct 2 '16 at 18:15

When $n=3$, the group of rational solutions is given by the points $(x,y)\in \{(0,\pm 1), (1/2,\pm 3),(-1/4,0)\}$, along with the point at infinity. The only $x$ value among these points which is a (proper) perfect power is $0$, so the only solutions when $3|n$ and $n>3$ are the trivial ones.

When $n=4$, we'll assume joro's Sage computation, that there are no non-trivial rational points. This also deals with the case when $4|n$.

Now, we may as well replace $n$ by its largest prime factor, so we reduce to dealing with the case when $n=p\geq 5$ is prime. We can quickly check that $y\neq 0$, and reduce to looking for solutions when $y>0$. Write $y=a/b$ with $a,b$ positive, relatively prime integers. Similarly, write $x=c/d$ with $c\in \mathbb{Z}-\{0\}$, $d>0$ an integer, and $\gcd(c,d)=1$. We then have $$a^2-b^2=2^6b^2\left(\frac{c}{d}\right)^p.$$ The LHS is an integer, hence so is the RHS. We will look at two main cases.

Case 1: $b$ is odd.

For any prime $q|b$, the LHS has $q$-adic valuation $0$, hence so does the RHS. So all of the primes in $b$ are accounted for in $d$. There are thus two cases, depending on whether $2$ divides $d$:

Case 1a: Say $d$ is odd. Then $d^p=b^2$. Thus, $b=s^p$, $d=s^2$, for some odd integer $s>0$. Then we have $$(a-s^p)(a+s^p)=2^6c^p.$$ The two factors on the left hand side have GCD at most $2$. On the other hand, at least one of the factors must in fact be divisible by $2$. Hence either $$a-s^p=2u^p,\qquad a+s^p=2^5v^p$$ or $$a-s^p=2^5u^p,\qquad a+s^p=2v^p$$ where $c=uv$ with $\gcd(u,v)=1$. Solving for $a$, and plugging into the other, and dividing by $2$, we get $$(\ast)\qquad s^p+u^p=2^4v^p \qquad \text{ or } \qquad s^p + 2^4u^p = v^p.$$ Note that these are really the same equation, after moving all terms to one side, using the fact that we can bring $-1$ into a $p$th power.

Case 1b: $d$ is even. Then (by similar reasoning as in case 1a) $p=5$, $b=s^5$, $d=2s^2$, for some odd integer $s>0$. Then we have $$(a-s^5)(a+s^5)=2c^5.$$ Now, $a$ must be odd, so that $2$ divides the LHS. But then the LHS has at least $2$-adic valuation $\geq 2$, hence $2|c$, which contradicts $\gcd(c,d)=1$.

Case 2: $b$ is even. Say $b=2^{k}e$ for some odd integer $e>0$, and some integer $k\geq 1$.

By the same reasoning as in case 1, we must have that $d^p=2^{2k+6}e^2$, hence (since $p$ is odd) we get $k=pt-3$ for some integer $t\geq 1$, $d=2^{2t}s^2$, and $e=s^p$ for some odd integer $s>0$. Hence, we obtain $$(a-2^{pt-3}s^p)(a+2^{pt-3}s^p)=c^p$$ with $c$ odd. The factors on the LHS are relatively prime, and equal to a $p$-power, hence are both $p$-powers. Thus, solving as before, we eventually arrive at $$(\ast\ast)\qquad \frac{1}{4}w^p + u^p=v^p$$ where $w=2^ts$.

In other words, moving terms accordingly, you can reduce your equation to solving the two Fermat-like equations $$X^p+Y^p=16Z^p\qquad \text{ or } \qquad 4X^p+4Y^p=Z^p.$$ I'm not certain if the modularity methods for solving Fermat's last theorem apply here, but that's probably the next place to look.