Getting Down with Fermat — A Lesson in Infinite Descent
Pierre de Fermat, a seventeenth century French lawyer and mathematician needs no introduction, he is famous for the notoriously hard conjecture, Fermat’s Last Theorem(FLT), which took over three centuries and a half to prove after a series of failed attempts by cranks and professional mathematicians alike.
“It is impossible to separate a cube into two cubes, a fourth power into two fourth powers, or generally, any power above the second into two powers of the same degree”, Fermat wrote in the margin of his copy of an ancient math text. What that means in mathematical notation is that there are no integers x, y, z that satisfy the equation xⁿ + yⁿ = zⁿ when n is an integer greater than two. Fermat and other mathematicians were able to prove the conjecture for some specific values of n using the method of infinite descent also called Fermat’s method of descent. The proof of the general case remained elusive for a long time though Fermat claimed to have a proof which was too long to fit into the margin of his book. The only recorded proof we have from Fermat is that of the case n = 4 using his method of infinite descent.
The Pythagorean Prelude
The familiar theorem of Pythagoras; a² + b² = c², gives the relationship between the sides of a right-angled triangle, where c is the length of the hypotenuse, and a, b are the lengths of the other two sides.
For those who like to take the hypotenuse of life, Pythagoras theorem comes in handy in calculating the exact savings on distance. The ancient folks also found it useful in their civil constructions and were naturally interested in when the lengths of the sides of a right-angled triangle will have integer values. There are ancient records containing sets of three integers that satisfy the Pythagoras equation like (3, 4, 5), (5, 12, 13), (8, 15, 17), (7, 24, 25), (20, 21, 29), to list a few. These are called Pythagorean triples. For example, plugging in the the triple (3, 4, 5), we have 3² + 4² = 5² (9 + 16 = 25).
What is even more interesting is the formula for generating such triples found in the ancient book, Euclid’s Elements. The formula states that the expressions a = p²-q², b = 2pq, c = p²+q² form a Pythagorean triple, where p and q are arbitrary integers greater than 0, and p is greater than q. We can verify the correctness of this formula by substituting the expressions for a, b, c in the Pythagoras equation as follows:
a² + b²
= (p²-q²)² + (2pq)²
=(p⁴-2p²q²+q⁴) + 4p²q²
=2p²q² + p⁴ + q⁴
=2p²q² + p⁴ + q⁴
So, from the above substitutions and expansions we see that the triple generating formula agrees with the Pythagoras theorem and therefore correct. For an example, let p=6, q=4, this yields the triple (20, 48, 52). Each number in this triple is a multiple of 4, we can factor it out to get the smaller triple (5, 12, 13), a triple in this reduced form with no common factor is called a primitive triple. To generate a primitive triple from the start, make sure the p and q are relatively prime, that is, they have no common factor, and also of opposite parity; one odd, the other even. We need all this to prove the n= 4 case of Fermat’s Last theorem in the next section.
Proof of FLT for exponent 4
A basic fact : if FLT is true for an exponent n, then it is also true for all the multiples of that exponent, that is, if xⁿ + yⁿ has no integer solutions so does xᵐ + yᵐ, where m=kn, a multiple of n, since xᵐ + yᵐ can be written as
(xᵏ)ⁿ + (yᵏ)ⁿ which is identical to the original equation. Now, every number greater than 2 is either a multiple of 4 or a multiple of some prime number p. Consequently, proving FLT reduces to proving the case n = 4 and the case n = p, where p is an arbitrary prime number. So, the proof of case n= 4 is quite significant, though very simple. The n=p case was never (fully) proved. FLT was proved via a completely different approach.
We want to prove that x⁴ + y⁴ = z⁴ has no solutions in positive integers using Fermat’s method of infinite descent. We do this by proving the stronger statement that x⁴ + y⁴ = z² has no solutions in positive integers. This is kind of easier to prove and takes care of the original equation since x⁴ + y⁴ = z⁴ can be rewritten as x⁴ + y⁴ = (z²)².
Step 1. x⁴ + y⁴ = z² can be rewritten as (x²)² + (y²)² = z² — this looks like Pythagoras theorem with x², y², z as the lengths of the sides of the triangle. Let’s assume this equation has solutions in positive integers(x, y, z), then (x², y², z) is a Pythagorean triple, so we have x²=2pq, y² =p²-q², z = p²+q², where p, q are some positive integers. Note that x² is a multiple of 2, it is even, this implies x is even too.
Step 2. The equation y² =p²-q² from step 1 can be rewritten as y² + q² =p², this also looks like Pythagoras theorem, and since y, p, q are all integers, they are a Pythagorean triple, so we have q = 2rs, y =r²-s², p =r²+s², where as usual, r and s are arbitrary positive integers. We can assume that (y, p, q) is a primitive triple, this implies that r and s are relatively prime.
Step 3. From step 2, we have q = 2rs, p =r²+s², substitute these values of q and p into x²=2pq from step 1; x²=2(r²+s²)(2rs) = 4rs(r²+s²),
x²/4 = rs(r²+s²)
(x/2)² = rs(r²+s²)
Since x is even, x/2 is an integer, let’s call it k, so we have:
k² = rs(r²+s²)
Now, since r and s are relatively prime, so are rs and r²+s². And since the product rs(r²+s²) equals a perfect square(k²), rs and (r²+s²) must be perfect squares too(check proof here). By the same argument, r and s are individually perfect squares. So there exist some integers u, v, w, such that r = u², s=v², r²+s² = w².
Substitute r = u², s=v² into r²+s² = w² to have :
u⁴+v⁴ = w²
Observe that the equation we just obtained; u⁴+v⁴ = w², is similar to the equation we started with; x⁴ + y⁴ = z².
If (x, y, z) is a solution of the original equation like we assumed, then (u, v, w) is another solution we obtained by algebraic manipulations. To complete our descent argument, we need to show that this second solution is smaller than the first, we do that in step 4.
Step 4. From step 2, we have p = r²+s². Obviously, p is less than p²+q², that means r²+s² is less than p²+q².
But from step 3, r²+s² = w². And from step 1, p²+q² = z. Therefore, from the preceding statement, w² is less than z, which obviously implies w is less than z. So (u, v, w) is a smaller solution set than (x, y, z). This a descent on z.
We could repeat the above steps on the new equation u⁴+v⁴ = w², since it has the same form as the original equation, and get a still smaller solution set and we can go on repeating the steps ad infinitum to get smaller and smaller solutions. This infinite descent contradicts the well-ordering principle; there is a least positive integer, the number 1, infinite descent implies we can go lower than the number 1, which is not true. So the initial assumption that x⁴ + y⁴ = z² has solution in positive integers is false. Proof completed.
x⁴ + y⁴ = z² is an example of a Diophantine equation, whose solution is required to be integers, and we’ve just shown that such solution does not exist using the method of infinite descent. The method of infinite descent is very useful for proving that solutions do not exist for certain Diophantine equations. To consolidate our understanding of the technique, let’s take another example.
Does the equation x² + y² = 3z² have integer solutions? Let’s find out.
Step 1. Assume there is a solution in integers: (x, y, z). Where z is not 0.
Step 2. Since z is assumed to be a non-zero integer in step 1, that means
x² + y² is a multiple of 3 according to the equation in question. This implies x and y are individually multiples of 3 too, so we can write x = 3m, y = 3n, where m and n are arbitrary integers.
Step 3. Substitute x = 3m, y = 3n into the original equation to get:
9m² + 9n² = 3z².
Divide the above equation by 3, a common factor of the terms, to get:
3m² + 3n² = z² ⇨ 3(m² + n²) = z²
Step 4. z² =3(m² + n²) from step 3, and m, n are integers, this means z² is a multiple of 3, and since 3 is a prime number, z must be a multiple of 3 too. So we can write z = 3k, where k is an arbitrary integer.
Now, substitute z = 3k in the equation 3(m² + n²) = z² to get:
3(m² + n²) =9k²
Divide the above equation by 3, the common factor to get:
m² + n² =3k².
Observe that the equation we just obtained, m² + n² =3k², is identical to the equation we started with, that means (m, n, k) is another solution to the equation, with k less than z(remember z = 3k, and k an integer). We can repeat the same steps for the later equation to get a yet smaller solution and go on ad infinitum. So, by the Infinite Descent argument, it is wrong to have assumed an integer solution. Therefore, the equation x² + y² = 3z² has no integer solutions.