An ancient result states that a triangle with vertices , and with lengths , and is right angled at iﬀ . The fact that is a necessary condition for the right angle is the famous Pythagoras theorem. However, it is also a suﬃcient condition for the triangle to be a right angle. In fact, it is known that
- a triangle is obtuse at iﬀ ;
- a triangle is right angles at iﬀ ;
- a triangle is acute at iﬀ .
In all the above cases the suﬃciency can be proved using the law of cosines. An important application of above result is that it can used to construct right angles (say in a construction of a building).
What are all the integer triples for which ? For instance, is not a integer triple that solves . Note that if a integer triple solves then all combinations of are also integer triple. Therefore, it is enough to look for non-negative integer triples . Further, and is also a trivial integer solution which will not represent a triangle. A positive integer triple is said to be Pythagorean triple if . Note that if is a Pythagorean triple then is also a Pythagorean triple for all . For instance, is a Pythagorean triple then (multiplied by ) is also a Pythagorean triple. This corresponds to the case similar right triangles. A Pythagorean triple is said to be primitve if , i.e., have no common factors, except , among them. This is equivalent to saying that because whatever divides LHS will also divide RHS. For instance, and are both primitive Pythagorean triple (PPT).
As we have already noted, there are integer triples that are not PPT. For instance, is not a PPT. Therefore, two natural questions are:
- How many PPT’s are there, ﬁnite or inﬁnite?
- Is it possible to generate all PPT’s?
It turns out that there are inﬁnitely many PPT’s and the answer to second question is in aﬃrmative.
Proof. Since both cannot be even. Suppose both and and are odd, then and . Therefore and are also odd because square of an odd number is odd (Use !). Therefore is even because sum of odd numbers is even. Thus is even and, hence is even. Let . Then
Note that LHS is odd and RHS is even which is a contradiction. Therefore, one of and is even and the other odd. This implies is odd and, hence, is odd. □
The above result rules out all triples , with even, as a possible PPT. It is enough to consider odd. But that still does not say all odd are allowed. For instance, there is no choice of such that is a PPT. In fact, the ﬁrst PPT is .
Also owing to the result above, henceforth in the pair , we shall always denote the odd number by and the even number by . For any PPT , we write . Note that .
Proof. Let divide both and . Then and for some . Then and which means that divides both and . But and, hence, or . Since should also divide , hence . But is odd so . □
Above result says that , a perfect square, is product of two relatively prime numbers and . Let and be the respective unique prime decomposition. Then . Since and are relatively prime for all . Thus, each and is even. Therefore, and are also perfect squares, say and , for some such that and . Consequently, and, since is odd, both and are odd. So, for every choice of odd positive integer such that and , we have a PPT given by , and . The last two equalities are obtained by solving for and using and . Since there are inﬁnitely many choices of satisfying above condition there are inﬁnitely many PPT’s.
A generalisation of the Pythagorean triple condition is, for a ﬁxed integer , seeking triples such that . Obviously, there are some trivial solution if is such that . However, it turns out there are no non-trivial solution, i.e., with
Suppose . Then if the equation has integer solution then, using , is a solution corresponding to . Therefore, to prove FLT it is enough to prove the result for . Note that any is:
- either , for some integer , which is same as . Choose in this case;
- or an odd prime. Choose in this case;
- or a muliple of an odd prime. Choose to be the odd prime.
Thus, it is enough to prove the result for and odd prime . If there is an integer solution to then is an integer solution to . So, to prove FLT for we show the following theorem:
Proof. Suppose there is a integer triple such that solving . Without loss of generality, we may assume that are all positive, and, hence, .
(Step 1): Note that is a PPT, since (if necessary, we rewrite and such that is odd and is even). We also know that is always odd.
(Step 2): There exists odd numbers , such that , and
Then, . Since and are odd, both and are even and, hence, are not coprime.
(Step 3): Let divide both and . Then and for some . Then and which means that divides both and . But therefore divides . Therefore, . Hence, , , and . Since , .
(Step 4): In fact, using the equation of , we have , i.e., is a PPT. Therefore, is even and is odd. Let . Then .
(Step 5): Using equation for , we get , i.e., . Hence and are perfect squares, and and . Therefore, is a PPT.
We repeat Step 2 on . There exists odd numbers , such that and we have
We repeat Step 3 on and to obtain and such that , , , and .
We repeat Step 5 on to conclude that and, hence, and and .
Using the value of and in the equation of , we get
Therefore, is also a non-trivial integer solution of . Let us compare the two solutions and . Note that
This implies that , hence . Thus, when we started with a non-trivial solution we obtained another non-trivial solution such that . We can repeat the entire proof for again for playing the role of above and can obtain another non-trivial solution such that . This is a contradiction because we cannot carry on inﬁnite number of times as suggested by our proof. Therefore, our assumption on the existence of a non-trivial integer solution is false. □
With all the eﬀort above, it only remains to prove the Fermat’s Last theorem for odd primes.
The odd primes are all regular primes. The ﬁrst irregular prime is . It is an open question: are there inﬁnitely many regular primes? However, it is known that there are inﬁnitely many irregular primes.
Gerhard Frey proved the following result related to FLT:
then the following elliptic curve, called Frey’s curve, must exist:
So, what is an elliptic curve? Given , consider plane curve of the form . Let us call it . The discriminant of is deﬁned as
The relates to the case when the curve self-intersects, called singular points. A curve is said to be an elliptic curve if . If the curve will divide the plane in two connected components and if it divides the plane into more than two connected components. Elliptic curves are symmetrical about -axis. The points on the elliptic curve form an abelian group under a suitable binary operation (). An elliptic curve is modular if there exists a -function of . In 1955, the Taniyama-Shimura-Weil (TSW) conjectured (also called modularity theorem) states that:
This means if Frey’s curve existed it must be modular. But, in , Jean-Pierre Serre conjectured (epsilon conjecture) that Frey’s curve is not modular which was later proved by Ken Ribet in . Since Frey’s curve is an elliptic curve it must be modular assuming TSW conjecture. An elliptic curve is semistable if a prime divides its discriminant and Frey’s curve is semistable. In - by Andrew Wiles showed that:
This means Frey’s curve cannot exist and, hence, Fermat’s last theorem is true.