# infinite descent

Fermat invented this method of infinite descent. The idea is: If a given natural number^{} $n$
with certain properties implies that there exists a smaller one with these properties, then
there are infinitely many of these, which is impossible.

Here is an example:

Let $m,n$ be coprime^{} positive integers with opposite parity, $$, and, say, $m$ is even.

Let $a=2mn$, $b={n}^{2}-{m}^{2}$, $c={m}^{2}+{n}^{2}$. Then $\{a,b,c\}$ is a primitive Pythagorean triple^{},
and the area $A$ of the right triangle^{} with sides $a,b,c$ is $ab/2=mn({n}^{2}-{m}^{2})$.

Suppose $A$ is a square. Then, since $m,n$ are coprime and of opposite parity, $\mathrm{gcd}(m+n,m-n)=\mathrm{gcd}(m,n)=1$. Thus, for $A$ to be a square, each of $m,n,m-n,m+n$ must be squares itself. Setting ${r}^{2}=m$, ${s}^{2}=n$, we have $A={(rs)}^{2}({s}^{4}-{r}^{4})$.

We prove that the Diophantine equation^{} ${x}^{4}-{y}^{4}={z}^{2}$ has no solution in natural numbers.

###### Remark 1.

Suppose that ${z}^{\mathrm{2}}\mathrm{+}{y}^{\mathrm{4}}\mathrm{=}{x}^{\mathrm{4}}$, where $\mathrm{gcd}\mathit{}\mathrm{(}x\mathrm{,}y\mathrm{,}z\mathrm{)}\mathrm{=}\mathrm{1}$, $x\mathrm{,}y\mathrm{,}z\mathrm{\in}\mathrm{N}$. Then $x$ is odd, and $y\mathrm{,}z$ have opposite parity.

###### Proof.

If $x$ was even, then ${x}^{4}={z}^{2}+{y}^{4}\equiv {(z+{y}^{2})}^{2}\equiv 0\phantom{\rule{veryverythickmathspace}{0ex}}(mod2)$, so $z,{y}^{2}\equiv 0\phantom{\rule{veryverythickmathspace}{0ex}}(mod2)$ or $z,{y}^{2}\equiv 1\phantom{\rule{veryverythickmathspace}{0ex}}(mod2)$. But $z,{y}^{2}\equiv 0\phantom{\rule{veryverythickmathspace}{0ex}}(mod2)$ conflicts with $\mathrm{gcd}(x,y,z)=1$. And $z,{y}^{2}\equiv 1\phantom{\rule{veryverythickmathspace}{0ex}}(mod2)$ implies ${y}^{2}+{({z}^{2})}^{2}\equiv 2\phantom{\rule{veryverythickmathspace}{0ex}}(mod4)$ contradicting ${x}^{4}\equiv 0\phantom{\rule{veryverythickmathspace}{0ex}}(mod4)$. Thus, $x$ is odd, and ${x}^{4}={z}^{2}+{({y}^{2})}^{2}\equiv {(z+{y}^{2})}^{2}\equiv 1\phantom{\rule{veryverythickmathspace}{0ex}}(mod2)$ implies that $z,{y}^{2}$ have opposite parity. ∎

Suppose $x$ is odd and $z$ is even. Then we have $z=2pq$, ${y}^{2}={q}^{2}-{p}^{2}$ and ${x}^{2}={q}^{2}+{p}^{2}$, where $p,q$ have opposite parity and are coprime. Since $z$ is odd, this implies ${(xy)}^{2}={q}^{4}-{p}^{4}$, so it is sufficient to show that there is no solution for odd $z$.

Now $x,z$ are assumed odd. Then $y$ is even, and there exist $m,n\in \mathbb{N}$, $$,$(2mn,m+n)=1$ such that

${y}^{2}$ | $=2mn$ | (1) | |||

${x}^{2}$ | $={n}^{2}$ | $+{m}^{2}$ | (2) | ||

$z$ | $={n}^{2}$ | $-{m}^{2}.$ | (3) |

Since ${m}^{2}+{n}^{2}={x}^{2}$ is a primitive Pythagorean triple, there exist $p,q\in \mathbb{N}$, $$, $(2pq,p+q)=1$ satisfiying

$m$ | $=2pq$ | (4) | |||

$n$ | $={q}^{2}$ | $-{p}^{2}$ | (5) | ||

$x$ | $={q}^{2}+{p}^{2}.$ | (6) |

Since $2mn$ is a square and $m,n$ are coprime and, say, $n$ is odd, $n$ is a square, and we have $m=2{r}^{2}$, $n={s}^{2}$.

From the primitive Pythagorean triple ${m}^{2}+{n}^{2}={x}^{2}$ we get $x={u}^{2}+{v}^{2}$, $n={u}^{2}-{v}^{2}$, $m=2uv$. Since $2uv=2{r}^{2}$ $uv$ is a square, and each of $u$ and $v$ is a square: $u={g}^{2}$, $v={h}^{2}$.

Substituting $n,u,v$ in $n={u}^{2}-{v}^{2}$ we have ${s}^{2}={g}^{4}-{h}^{4}$. But since $$ this implies $$, thus we have another solution with odd $$. This contradicts to the fact that there exists a smallest solution.

See http://mathpages.com/home/kmath144.htmhere for a discussion of
infinite descent vs. induction^{}.

Title | infinite descent |
---|---|

Canonical name | InfiniteDescent |

Date of creation | 2013-03-22 14:07:56 |

Last modified on | 2013-03-22 14:07:56 |

Owner | Thomas Heye (1234) |

Last modified by | Thomas Heye (1234) |

Numerical id | 13 |

Author | Thomas Heye (1234) |

Entry type | Topic |

Classification | msc 11D25 |

Related topic | ExampleOfFermatsLastTheorem |