# 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 $x\geq-1$. 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 $x\equiv-1\pmod{d}$, so that $1+x^{2}\equiv 2\pmod{d}$. But $d\mid 1+x^{2}$ so that $2\equiv 0\pmod{d}$ 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

 $\displaystyle 1+x=2r^{2}$ $\displaystyle 1+x^{2}=2s^{2}$

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

 $\displaystyle r^{2}=p^{2}-q^{2}$ $\displaystyle r^{2}-1=2pq$ $\displaystyle s=p^{2}+q^{2}$

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

 $\displaystyle r^{2}=2pq$ $\displaystyle r^{2}-1=p^{2}-q^{2}$ $\displaystyle s=p^{2}+q^{2}$

From the second of these formulas, we see that $p$ must be even (consider both sides $\pmod{4}$), say $p=2t^{2}$. Now,

 $\displaystyle(p+q-1)(p+q+1)=(p+q)^{2}-1=p^{2}+2pq+q^{2}-1=\\ \displaystyle p^{2}+2pq+q^{2}-(r^{2}-(r^{2}-1))=p^{2}+2pq+q^{2}-2pq+p^{2}-q^{2% }=2p^{2}=8t^{4}$

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,

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

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 $1+x+x^{2}+x^{3}=y^{2}$ SolutionsOf1xx2x3y2 2013-03-22 17:05:09 2013-03-22 17:05:09 rm50 (10146) rm50 (10146) 10 rm50 (10146) Theorem msc 11F80 msc 14H52 msc 11D41