Nice Post-Dinner Diophantine – Mattia Giuri's bizarre blog

Today after dinner, I really wanted to do a geometry problem but this diophantine equation just suddenly appeared in front of my eyes and, because it’s been a while since the last number theory problem, I decided to give it a shot.

Prove that

$x^3 + 3 = 4y(y+1)$

has no integer solutions.

First we notice that the former equation is equivalent to

$x^3 + 4 = (2y+1)^2$

$\Rightarrow x^3 = (2y+3)(2y-1)$

Now we notice that $gcd(2y+3, 2y-1) = gcd(2y-1, 4) = 1$ , meaning that $2y+3 = u^3, 2y-1 = v^3$ .

But no cubes differ by 4, implying there is actually no integral solution to such equation.

That was a pretty straightforward problem, but we can deduce that the best way of finding the solutions (or proving they don’t exist) of similar diophantines is trying to avoid sums and only have products on both sides of the equation, and that requires good manipulation abilities which only come with practice.