This question is simplified case of
Integer solutions of $a^6 + 4 b^3 = c^6$.
Is there integer solutions for $$a^6 + 4 b^3 = 1$$
The way I am trying to prove is based on Fermat's Theorem, but I can not get the final result.
Asked
Active
Viewed 101 times
2
-
1We must have $a\equiv 1,3 \pmod4$ and $(a,b)=1$ – Dan Jan 05 '18 at 11:19
1 Answers
6
From $a^6+4b^3=1$, we see that $a$ must be odd. Let's rewrite the equation as
$$(a^3-1)(a^3+1)=-4b^3$$
Since $a^3$ is odd, we have $\gcd(a^3-1,a^3+1)=2$, so we have
$$\begin{align} a^3-1&=2c^3\\ a^3+1&=2d^3 \end{align}$$
where $cd=-b$ with $\gcd(c,d)=1$. But this implies
$$a^3=c^3+d^3$$
which, as Fermat observed, has no nontrivial integer solutions. So we're left with $(a,b)=(\pm1,0)$ as the only integer solution.
Remark (added later): As the OP brilliantly observes, there is really no need to rely on Fermat. If you subtract the two equations instead of adding them, you get $d^3-c^3=1$, which immediately eliminates anything nontrivial, since cubes have a hard time differing by $1$.
Barry Cipra
- 81,321
-
I think would be simpler just to observe the difference $d^3-c^3=1$ which is impossible. So no need to refer to Fermat theorem – Gevorg Hmayakyan Jan 05 '18 at 15:14
-
1
-
Actually this proof can be generalized for any $p$ : $a^{2p}+4b^p=1$ has no integer solutions for any $p$. – Gevorg Hmayakyan Jan 05 '18 at 16:18