# Solve the diophanic equation $y^2 = x^4 -4x^3 + 6x^2 -2x +5$, where $x,y \in \mathbb{Z}$

Solve the diophanic equation $y^2 = x^4 -4x^3 + 6x^2 -2x +5$.

Methods I know:

1) look modulo p for some prime p, when using this method I almost always conclude there are no solutions, so i don't think it is handy to use this in this particular case.

2) Combinations of factorization and estimation: factorize your specific function and Search for an upper bound.

I used 2), and i get in a little bit of trouble.

I tried factorize the right hand side, but did not come any further then:

$y^2 - 10 = (x-5)(x^3 + x^2 - x - 3)$.

I don't seem to get an upper bound here..... :(.

Any hints on going to the right direction or on using a new method trying to solve this probelem?

Kees

• WA gives solutions, but no explanation (link). I wonder if it just does a search for small numbers. – mvw Nov 17 '15 at 14:17

We may suppose that $y\geq 0$. Now $$x^4-4x^3+6x^2-2x+5=(x-1)^4+2x+4$$ Hence we have $$(y-(x-1)^2)(y+(x-1)^2)=2(x+2)$$
This imply that $(x-1)^2\leq (x-1)^2+y\leq 2|x|+4$ and it is easy to finish.
• wow, nice factorization you got there. Why can we assume that $y \geq 0$ ? – Kees Til Nov 17 '15 at 14:26
• If $(x,y)$ is a solution, then $(x,-y)$ is also a solution, so we can suppose that $y\geq 0$ to bound $|x|$. – Kelenner Nov 17 '15 at 14:28
• hmmmmm, true that. Are we trying to find a bound of $|x|$ in the last step, because i don't see how |x| cant go to infinity if we have: $|x| \leq - \frac{(x-1)^2 - 4}{2}$ – Kees Til Nov 17 '15 at 14:45
• We have $x^2-2|x|+1\leq x^2-2x+1\leq 2|x|+4$, so $x^2-4|x|-3\leq 0$, etc – Kelenner Nov 17 '15 at 14:49