Pythagorean Triples and Fermat's Last Theorem

The Pythagorean Theorem says that the sum of the squares of the sides of a right triangle equals the square of the hypotenuse. In symbols,

${a}^{2}+{b}^{2}={c}^{2}$.

Here are a few examples which satisfy the Pythagorean Theorem:

${3}^{2}+{4}^{2}={5}^{2}$; ${7}^{2}+{24}^{2}={25}^{2}$; ${8}^{2}+{15}^{2}={17}^{2}$; ${65}^{2}+{72}^{2}={97}^{2}$.

Are there infinitely many Pythagorean triples? The answer is "YES" for a trivial reason since, for example, for any integer d,

${\left(3d\right)}^{2}+{\left(4d\right)}^{2}={\left(5d\right)}^{2}$

These new triples are not interesting, so we concentrate only on triples with no common factors. Such triples are called primitive Pythagorean triples.

Here are some other triples:

(20; 21; 29); (12; 35; 37); (9; 40; 41); (16; 63; 65); (28; 45; 53).

Pythagorean Triples Theorem:

 You will get every primitive Pythagorean triple (a; b; c) with a odd and b even by using the formulas: $a=st$,  $b=\frac{{s}^{2}-t{}^{2}}{2}$, $c=\frac{{s}^{2}+t{}^{2}}{2}$ where s > t ≥1 are chosen to be any odd integers with no common factors.

Alternatively, since b is even, we could have started with ${b}^{2}={c}^{2}-{a}^{2}$. Hence ${\left(\frac{b}{2}\right)}^{2}=\frac{c-a}{2}.\frac{c+a}{2}$ and, we conclude that $\frac{c+a}{2}={u}^{2}$ and $\frac{c-a}{2}={v}^{2}$ so that $c={u}^{2}+{v}^{2}$$a={u}^{2}-{v}^{2}$ and b = 2uv where u > v ≥ 1, (u,v) = 1, and u and v have opposite parity.

Since the equation ${a}^{2}+{b}^{2}={c}^{2}$ has infinitely many solutions, it is natural

to investigate the situation where the exponent 2 is replaced by 3, and then 4, and so on.

For example, do the equations

${a}^{3}+{b}^{3}={c}^{3}$, ${a}^{4}+{b}^{4}={c}^{4}$, and ${a}^{5}+{b}^{5}={c}^{5}$

have solutions in nonzero integers a, b, c?

In 1637 Pierre de Fermat(1601-1665) showed that there is no solution for exponent 4.

Theorem: Fermat(1637)

 ${a}^{4}+{b}^{4}={c}^{4}$ has no nontrivial solutions in integers.

Proof:

Assume that ${a}^{4}+{b}^{4}={c}^{2}$ where (a, b, c) have no common factor and say b is even. Then,

${b}^{2}=2uv$

${a}^{2}={u}^{2}-{v}^{2}$

$c={u}^{2}+{v}^{2}$

where u > v ≥1, (u, v) = 1; and u and v of opposite parity. In fact, v must be even for if not and u is even, then from ${a}^{2}={u}^{2}-{v}^{2}$, we have ${a}^{2}\equiv -1\equiv 3$ (mod4) which is not possible. Since ${a}^{2}+{v}^{2}={u}^{2}$, then as above

$v=2pq$

$a={p}^{2}-{q}^{2}$

$u={p}^{2}+{q}^{2}$

Therefore ${b}^{2}=2uv=4pq\left({p}^{2}+{q}^{2}\right)$ and hence p, q, pq and ${p}^{2}+{q}^{2}$ are squares since p, q, pq and ${p}^{2}+{q}^{2}$ have no factors in common. Let ${p}^{2}={A}^{2}$ and $q={B}^{2}$. Then ${A}^{4}+{B}^{4}={p}^{2}+{q}^{2}$ is a square and

${A}^{4}+{B}^{4}={p}^{2}+{q}^{2}$ = u < ${u}^{2}+{v}^{2}=c$ < ${c}^{2}={a}^{4}+{b}^{4}$

This sets up an infinite descent chain of squares of whole numbers of the form ${x}^{4}+{y}^{4}$ which is clearly impossible. (The above argument is called the method of infinite descent and was invented by Fermat.)