Fermat's lost theorem

Summary of article : Fermat's lost last theorem : can consecutive squares account for the intuition behind Fermat's conjecture?

Did Pierre Fermat really have proof of his conjecture ?

If not, what information did he have when studying Pythagorean triples?

Fermat’s Last Theorem is the assertion, that x^n + y^n = z^n (1) has no solution in positive integers when n > 2 and x, y, z # 0. Consequently only Pythagorean triples provide whole number solutions to the equation when n = 2. The initial Fermat conjecture has been shown to be true by Andrew Wiles of Princeton University [1] in 1995, using algebraic geometry, unknown at the time of Fermat: proving the Taniyama-Shimura conjecture concerning elliptic curves. Our endeavour consists in exploring methods Fermat could have used when he asserted on Bachet’s Diaphantus, Book II, problem 8 [2], that “On the one hand it is impossible for a cube to be written as a sum of two cubes or a fourth power to be written as a sum of two fourth powers, or for any number, which is a power greater than the second to be written as the sum of two like powers. I have a truly marvellous demonstration of this proposition, which this margin is too narrow to contain.”

Is it possible to prove the last theorem by studying and analysing the underlying equations and networks related to triples? And what role do consecutive squares play in the building of natural numbers?

If we look at the equation z² = x² + y² (1), and change the order, y² = z² - x² , we find that z (odd) and x (even) are consecutive natural numbers. For example 7² = 13² – 12². We shall call z the major odd number, y the minor odd, and x the even number, with the triple of integers {z, x, y} throughout this article. We can say that y can be any natural odd number greater than 1; nevertheless z, however, has a special characteristic: z – 1 is divisible by 4.

First, how do we find triples? If we start with y (any odd number > 1) all we need to do is square the number, add 1 and divide by 2. Thus z = ½(y² + 1) and x = ½(y² - 1). In other words, we have two consecutive whole numbers. So we can say that y² = z² – x². This shows us that y² equals the difference between two consecutive squares. However, it must be stressed that every odd number apart from 1 equals the difference between two consecutive square numbers: 3 = 2² - 1, 5 = 3² – 2², and so on. So we can show that any odd number y is such that y = a² – b² = (a – b)(a + b) = a + b = 2a -1. This gives us a method for finding a and b. Take any odd and add 1 to it, then divide by 2, and we have a. Thus a = ½(y + 1). And of course b = a – 1. We can now reduce y to an equation with two variables, y and a: y = 2a – 1 for any odd number except 1.

Here is an example: If y = 21, a = 11, and b = 10. So y = 11² – 10². Let’s now square y:

y² = 21² = 441 = 221 + 220. This gives us the triple {221, 220, 21} or we can express this as

441 = 221² - 220².

We can take any power of y, yⁿ = a² – b². We can square this value, (yⁿ)², and find triples. We can have the triple {z, x, yⁿ.}. Here is an example. Let y = 3, yⁿ = 27. (yⁿ)² = 729 = 365 + 364.

The triple then is {365, 364, 27}. We now see how triples can be formed by squaring any odd number, including powers of any odd.

We may wonder why z – 1 is divisible by 4. We use the equation z = ½(y² + 1) and subtract 1. So z – 1 = ½(y² + 1) – 1 = 1/2(y² – 1) = 1/2(y – 1)(y + 1). We can now see that (y – 1)(y + 1) is divisible by 8. As y is odd, (y – 1) and (y + 1) are even, one is divisible by 2, the other by 4. If for example y = 5, y – 1 = 4, y + 1 = 6. Consequently z – 1 is always divisible by 4.

For even numbers divisible by 4, x = c² – d² = (c – d)( c + d) = 2(c + d) = 4c – 4. As all powers of x greater than n = 1 can be divided by 4, xⁿ = 4c – 4, the difference of two consecutive odd squares or even squares. For instance, 12 = 2(4 + 2) = 4² – 2² and 10² = 26² – 24².

We may wonder why odd numbers are the differences between two consecutive squares. So let us look deeper into how squares correspond to the addition line of odd numbers.

A square is the sum of a continuous addition process of odd numbers: 1 + 3 = 4, 1 + 3 + 5 = 9, 1 + 3 + 5 + 7 = 16 and so on always starting with a. We shall now see why y = a² – b² using the addition line.

We first have 9 = 1 + 3 + 5. If we take away 1 + 3 we are left with 5. And we can notice that 5 = 9 – 4 = 3² – 2². This is true for all odd numbers. If we want to find a and b using the odd number addition line, we choose a number, say 11. Find when it appears along the line:

1 + 3 + 5+ 7 + 9 +11. Add all the numbers, 36 or 6². We now know that 11 = 6² – 5².

So how does this explain Fermat’s last theorem?

First argument: as yⁿ = a² – b², and as a = b + 1, both a and b cannot have the same n^th root.

If yⁿ = a² – b² then a² – b² = zⁿ – xⁿ. We can suppose that a² = zⁿ and b² = xⁿ. So n must equal 2 on the RHS although n can have any positive integer value on the LHS.

So equation (1) is not possible in integer terms for any odd number > 1. Thus yⁿ = a² – b² and this stands for every odd natural number. Whatever the exponent, an odd number always equals the difference between two consecutive squares. However, a can be a number raised to a power. For example we can take 31 = 16² – 15², and we notice that 16² = 2⁸. We can clearly see that 15 cannot have the same eighth root as it is differs from 16 by 1.

Second argument: the same applies to even numbers as c = d + 2, c and d cannot have the same n^th root. Let xⁿ = c² – d² then c² = zⁿ and d² = yⁿ, so n = 2 on the RHS of the equation.

Third argument: if zⁿ = xⁿ + yⁿ, then 2g – 1 = 2a – 1 + 4c – 4. where zⁿ = 2g – 1, yⁿ = 2a -1 and xⁿ = 4c – 4.

In simple terms, the equation becomes (g – a) – 2(c – 1) = 0.

What value of n satisfies this equation?

If we take 25 = 16 + 9, then g = 13, a = 5 and c = 5.

Thus (g – a) – 2(c – 2) = (13 – 5) – 2(5 – 1) = 0.

In other words, the equation is satisfied by the squares of triples, consequently the only solution can be given by n = 2.

There is much more to say about the way natural numbers generate triples. It seems that natural numbers are bound in several ways to consecutive squares, and so zⁿ = xⁿ + yⁿ cannot have integer solutions when n > 2, as Fermat supposed.

There are ways of checking the above arguments first by using the Euclid identity:

(p² + q²)² = (2pq)² + (p² – q²)² with p > q > 0 and p and q of opposite parity.

So z² = (p² + q²)² = 2g – 1, y² = (p² – q²)² = 2a – 1, and x = (2pq)² = 4c – 4.

If we know p and q, we can find g, a, and c. For p = 2, q = 1, g = 13, a = 5, c = 5.

If y² = (a² – b²)², then z² must equal (a² + b²)²_. If y = 5, y = 3² – 2². So z = 13. And we have the triple {13, 12, 5} or 13² = 12² + 5².

We can now assert that the equation for triples can be expressed with at least one of the odd or even members (x or y) as powers greater than 2. For instance 5² = 4² + 3² can be written

5² = 2⁴ + 3² so we can find z² = (xⁿ)² + y² or z² = x² + (yⁿ)².

If zⁿ = xⁿ + yⁿ, then (2g-1) = 4(c – 1) + (2a – 1) if we let zⁿ = 2g – 1, yⁿ = 2a – 1, and xⁿ = 4(c – 1).

When n =2, there are an infinite number of solutions.

However (2g-1) = (p² + q²)², (2a – 1) = (p² - q²)² and 4(c -1))= (2pq)². So when n = 2

(2g-1) = 2(c – 1) + (2a – 1) => (p² + q²)² = (2pq)² + (p² – q²)² with p > q > 0 and p and q of opposite parity.

But if n = 3 or n > 2 (p² + q²)³ # (2pq)³ + (p² – q²)³ and (p² + q²)ⁿ # (2pq)ⁿ + (p² – q²)ⁿ

This implies that (2g-1) # (2(c – 1)) + (2a – 1), with each term having the same nth root, consequently when n > 2, there can be no integer solution for xⁿ + yⁿ = zⁿ.

Finding g, a and c from the Euclid identity.

(2g-1) = (p² + q²)², so g = ½[(p² + q²)² +1]

(2a – 1) = (p² - q²)² so a = ½[(p² - q²)² + 1]

4(c -1))= (2pq)² so c = 1/4[(2pq)² + 4]

For p = 2, q =1, g = 13, a = 5, and c = 5.

We can now show why (zⁿ)² # (xⁿ)² + (yⁿ)²

We can now show why (zⁿ)² # (xⁿ)² + (yⁿ)².

Let y = uⁿ, where u and m are integers and m > 1. We know that z = ½(y² + 1). Substituting, we now have z = ½(u²ⁿ + 1). And we have x = ½(u²ⁿ - 1). We can clearly see that z and x cannot have the same n^th root as y. In other words, it is not possible to have

(zⁿ)² = (xⁿ)² + (yⁿ)².

We can now illustrate this last point by using the Pythagorean paradigm and adapting it to different values of n.

Let (z√z)^2 = (x√x)^2 + (y√y)^2 = x^3 + y^3 = z^3

We have three square roots to eliminate by substituting square values for z, x, and y. It is easy to eliminate two out of the three, but eliminating all three is impossible. This is why.

Let us assume that if all the terms of the set are to be integers at the same time then z must equal c^2, x = a^2, y = b^2. If we just make x and y squares, which is easy, we will always have a non integer z.

So we shall use the method of contradiction.

Suppose z^3 = x^3 + y^3 = (z√z)^2 = (x√x)^2 + (y√y)^2

Thus substituting c2 for z, a2 for x, b2 for y, to eliminate all the surds, we have

(c^3)^2 = (a^3) ^2 + (b^3) ^2,

c^6 = a^6 + b^6

If, x, y, z, are to be integers in this particular case, then a, b, c must be integers, but the equation c^6 = a^6 + b^6 must also be satisfied. Fermat showed that all even number greater than 2 could not have three integer solutions (his proof for n = 4). And we have shown the same conclusion above. So there are no solutions for equation 1 when n = 3. The same can be shown for n = 5 or n > 2.

Let v = (n-1)/2, then when n = 5, v = 2, n = 7, v = 3 and so on. Consequently

z^n = x^n + y^n = (z^v√z)² = (x^v√x)² + (y^v√y)²

Thus using the same method as before, (cn)² = (an)² + (bn)².

And so c^2n = a^2n + b^2n

Note that c < z, a < x, and b < y.

To conclude, we have noticed a very tightly knitted pattern relating natural numbers to consecutive squares, which seems to rule out the possibility of having integer solutions for the equation zⁿ = xⁿ + yⁿ except when n = 2 and the equation strictly corresponds to the Euclid identity. The main point is that any odd or even number, whatever the power, square, cube or greater, will always equal the difference between two consecutive squares or consecutive odd squares or even squares for even numbers.

Table 1: to show that g - a - 2(c -1) = 0