solutions of $1+x+x^2+x^3=y^2$

This article shows that the only solutions in integers to the equation

$$1+x+x^2+x^3=y^2$$

are the obvious trivial solutions $x=0,\pm 1$ together with $x=7,y=20$. This result was known to Fermat.

First, note that the equation is $(1+x)(1+x^2)=y^2$, so we have immediately that $x\ge -1$. So, noting the solutions for $x=0,\pm 1$, assume in what follows that $x>1$.

Let $d=\gcd (1+x,1+x^2)$. Then $x\equiv -1(modd)$, so that $1+x^2\equiv 2(modd)$. But $d\mid 1+x^2$ so that $2\equiv 0(modd)$ and $d$ is either $1$ or $2$. If $d=1$, so that $1+x$ and $1+x^2$ are coprime, then $1+x$ and $1+x^2$ must both be squares. But $1+x^2$ is not a square for $x>1$. Thus the $\gcd$ of $1+x$ and $1+x^2$ is $2$, so each must be twice a square, say

$$1+x=2r^2$$

$$1+x^2=2s^2$$

and then

$${(r^2)}^2+{(r^2-1)}^2={\left(\frac{1+x}{2}\right)}^2+{\left(\frac{-1+x}{2}\right)}^2=\frac{1+x^2}{2}=s^2$$

so that $r^2,r^2-1,s$ form a primitive Pythagorean triple (note that $r>1$ since $x>1$).

Recall that if $(a,b,c)$ is a primitive Pythagorean triple, then precisely one of $a$ and $b$ is even, and we can choose coprime integers $p,q$ such that $a=p^2-q^2,b=2pq,c=p^2+q^2$ or $a=2pq,b=p^2-q^2,c=p^2+q^2$ depending on the parity of $a$.

Suppose first that $r$ is odd. Then

$$r^2=p^2-q^2$$

$$r^2-1=2pq$$

$$s=p^2+q^2$$

Then $1=r^2-(r^2-1)={(p-q)}^2-2q^2$. Now, note that $p-q$ must be a square, say $p-q=t^2$, since $\gcd (p-q,p+q)=1$ and $(p-q)(p+q)$ is a square. Then

$$t^4={(p-q)}^2=2q^2+1={(q^2+1)}^2-q^4$$

so that

$$t^4+q^4={(q^2+1)}^2$$

But we know that the sum of two fourth powers can be a square (http://planetmath.org/ExampleOfFermatsLastTheorem) only for the trivial case where all are zero. So $r$ cannot be odd.

So suppose that $r$ is even. Then

$$r^2=2pq$$

$$r^2-1=p^2-q^2$$

$$s=p^2+q^2$$

From the second of these formulas, we see that $p$ must be even (consider both sides $(mod4)$), say $p=2t^2$. Now,

$$\begin{array}{c}(p+q-1)(p+q+1)={(p+q)}^2-1=p^2+2pq+q^2-1=\hfill \\ \hfill p^2+2pq+q^2-(r^2-(r^2-1))=p^2+2pq+q^2-2pq+p^2-q^2=2p^2=8t^4\end{array}$$

Since $p$ and $q$ have opposite parity, $p+q\pm 1$ are even, so that

$$2t^4=\frac{p+q-1}{2}\cdot \left(\frac{p+q-1}{2}+1\right)=u(u+1)$$

Thus one of $u,u+1$ is a fourth power and the other is twice a fourth power, say $b^4$ and $2c^4$.

Now,

$$u=b^4\Rightarrow u+1=2c^4\Rightarrow b^4-2c^4=-1$$

$$u=2c^4\Rightarrow u+1=b^4\Rightarrow b^4-2c^4=1$$

so that $b^4-2c^4=\pm 1$.

If $b^4-2c^4=1$, then

$${({(c^2)}^2+1)}^2={(c^2)}^4+b^4$$

and again we have a square being the sum of two fourth powers. So this case is impossible.

If $b^4-2c^4=-1$, write $e=c^2$, then ${(e^2-1)}^2=e^4-b^4$. It follows (see here (http://planetmath.org/X4Y4z2HasNoSolutionsInPositiveIntegers)) that either $b=0$ or $e^2-1=0$. If $b=0$, we get the impossibility $e^4-2e^2+1=e^4$, while if ${(e^2-1)}^2=0$, then $e^4=b^4,e=\pm 1$ and so $b=\pm 1$. Then ${\displaystyle \frac{p+q-1}{2}}=b^4=1$, so that $p+q=3$. Thus $r^2=4$, so $r=2$ and, finally, $1+x=2r^2=8$, so that $x=7$.

Thus, the only nontrivial solution to the equation given is

$$1+7+49+343=400$$

Title	solutions of $1+x+x^2+x^3=y^2$
Canonical name	SolutionsOf1xx2x3y2
Date of creation	2013-03-22 17:05:09
Last modified on	2013-03-22 17:05:09
Owner	rm50 (10146)
Last modified by	rm50 (10146)
Numerical id	10
Author	rm50 (10146)
Entry type	Theorem
Classification	msc 11F80
Classification	msc 14H52
Classification	msc 11D41