# 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 $d=\mathrm{gcd}(1+x,1+{x}^{2})$. Then $x\equiv -1\phantom{\rule{veryverythickmathspace}{0ex}}(modd)$, so that $1+{x}^{2}\equiv 2\phantom{\rule{veryverythickmathspace}{0ex}}(modd)$. But $d\mid 1+{x}^{2}$ so that $2\equiv 0\phantom{\rule{veryverythickmathspace}{0ex}}(modd)$ 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 $\mathrm{gcd}$ of $1+x$ and $1+{x}^{2}$ is $2$, so each must be twice a square, say

$$1+x=2{r}^{2}$$ | ||

$$1+{x}^{2}=2{s}^{2}$$ |

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

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

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 $p=2{t}^{2}$. Now,

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

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

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

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

Now,

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

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

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

If ${b}^{4}-2{c}^{4}=1$, 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 ${b}^{4}-2{c}^{4}=-1$, 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 ${e}^{4}-2{e}^{2}+1={e}^{4}$, while if ${({e}^{2}-1)}^{2}=0$, then ${e}^{4}={b}^{4},e=\pm 1$ and so $b=\pm 1$. Then $\frac{p+q-1}{2}}={b}^{4}=1$, so that $p+q=3$. Thus ${r}^{2}=4$, so $r=2$ and, finally, $1+x=2{r}^{2}=8$, 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 |