Complex Extensions, The xyn Lemma

x^j + yⁿ = zⁿ

Assume n and j are greater than 1. If j = n and n = 2, this is a pythagorean triple. If j = n and n > 2, this is a fermat triple.

Assume the variables are coprime. This may be a given, or, if n is a multiple of j, you can invoke the coprime lemma.

Let d = z-y, and replace z with y+d. Then pull yⁿ to the right hand side and write it this way.

x^j = (y+d)ⁿ - yⁿ

Expand (y+d)ⁿ by the binomial theorem. The terms yⁿ cancel, leaving something like this.

x^j = ndy^n-1 + (n:2)d²y^n-2 + (n:3)d³y^n-3 + … nd^n-1y + dⁿ

Let p be a prime. If p^k divides d, p^2k divides all but the first term on the right. If the first term, ndy^n-1, has fewer factors of p than all the others, then it sets the tone for the right side of the equation, and for x^j.

How many times does p divide ndy^n-1? By definition, p^k divides d. We know p does not divide y, else it would also divide z.

Assume p and n are coprime. The first term, ndy^n-1, is precisely divisible by p^k, not p^k+1. The same can be said for the entire right hand side, and for x^j. Therefore k is a multiple of j.

So far, d is a perfect j^th power.

Next assume p^r divides n, with p odd. Again, p^k divides d. We still want to show that ndy^n-1 sets the tone for the right hand side.

The first term has r+k powers of p, the second r+2k, and the third r+3k etc. Each additional instance of d introduces another k powers of p, but watch the binomial coefficients. We bring in n-1 on top and 2 on the bottom; then n-2 on top and 3 on the bottom etc. As a worst case, assume the numerator never brings in any additional factors of p. The ever expanding denominators pull one out at the p^th term, another at term 2p, another at 3p, and two factors are removed at term p², and so on. After w terms we have removed w/p + w/p² + w/p³ … factors of p. This is w times 1 less than the geometric series 1/p + 1/p² + 1/p³ etc. The series becomes 1/(1-1/p), or p/(p-1). Subtract 1 to get 1/(p-1). So the denominators remove at most w/(p-1) factors of p from the w^th term. Of course this is a worst case - an upper bound. Since p > 2, the denominators pull out at most ½w factors of p, while the powers of d bring in at least k(w-1) new factors of p. Even if k = 1, this is certainly larger than ½w. The powers of p keep growing, all the way to dⁿ.

When p = 3 and k = 1 we need to verify the second term by hand. The formula says we might be removing one factor of p, that is, 2/(3-1) = 1. And d might only bring in one factor of p, when k = 1. It looks like the second term and the first could both be divisible by p^k+r. In fact this can't happen, because the second denominator is 2, and that doesn't include any factors of 3. The second term, (n:2)d²y^n-2, has more factors of p than the first term, ndy^n-1, and subsequent terms have even more factors of p. The first term sets the tone for the entire sum, and k+r is a multiple of j.

This reasoning breaks down when p = 2; if d is even we can only hope n is odd, so that p and n are coprime.

Shared Primes

What if x y and z are not coprime? Let p divide all three variables. At the end of the coprime lemma, we proved that the same power of p divides yⁿ and zⁿ, hence the same power of p divides y and z. Call this p^r.

Now p^r, or something larger, divides into d. Assume p^r+k goes into d. In other words, d has more factors than y. Each term brings in another d, and takes away a y. Each term introduces another p^k. Yet this is precisely what we saw before. The first term, ndy^n-1, sets the tone for the right hand side, and for x^j. Relative to p, d is a j^th power, divided by the factors of p in ny^n-1.

It's not clear when this would be used, so return to the assumption that all variables are coprime.

n an odd prime

For the rest of this page, let n be an odd prime. Now p is either equal to n, or it is coprime to n. And (by assumption) p is coprime to y. Thus d, or z-y, is either a perfect j^th power, or it is a j^th power divided by n.

xⁿ + yⁿ = z^j

Since n is odd, we can show x+y is a j^th power, or a j^th power divided by n.

Let d = x+y, replace y with d-x, expand by the binomial theorem, and let the terms xⁿ cancel.

ndx^n-1 - (n:2)d²x^n-2 + (n:3)d³x^n-3 - … dⁿ = z^j

The alternating sum on the left is led by ndx^n-1, which sets the tone. Reasoning as above, d becomes a j^th power, or a j^th power divided by n.

Separating z

Continuing the above, let d = u^j, or perhaps u^j/n. The left side is divisible by d, and so is the right, which is z^j. If u^j divides z^j, then u divides z. (You can show this prime by prime.)

Let z = u×v, and suppose u and v have a prime factor p in common. Thus z^j, without u^j, has at least j factors of p. On the left side, pull d out. What remains has at least j factors of p. Since p divides d, it must also divide the first term, the only term without d, namely nx^n-1. If p divides x then x and z are not coprime. so p divides n, i.e. p = n. Now we can divide out another factor of p. Each binomial coefficient has one factor of p. And the last term, d^n-1, certainly has factors of p to spare. Now the first term is a power of x, and everything beyond this is still divisible by d. And we have at least j-1 factors of p to account for. Thus p divides x, which is a contradiction. Therefore u and v are coprime.

We know u^j = x+y, or x+y times n. Can we say anything about v?

Let p be a prime dividing v. Thus p divides xⁿ+yⁿ, but it does not divide x+y. In other words, xⁿ = -yⁿ = (-y)ⁿ mod p, but x is not equal to -y mod p. The ratio x over -y is not 1, but when raised to the n, we get 1. this means p-1 is a multiple of n, or if you prefer, p = 1 mod n. This holds for all primes in v, and v = 1 mod n.

xⁿ + yⁿ = zⁿ

In a fermat triple, where j = n, x+y, z-x, and z-y are all perfect n^th powers, or perhaps one of them is an n^th power divided by n. Hence I refer to this as the xyn lemma.

This gives you an idea of how difficult it is to build a fermat triple. The variables must yield n^th powers when added and subtracted, and beyond this, xⁿ+yⁿ is suppose to equal zⁿ. We're not surprised there are no such triples, although this does not constitute a proof.

Divisibility by 3 or 5

Start with the fermat triple xⁿ + yⁿ = zⁿ, where n is an odd prime. Let d = z-y. Replace z with y+d and expand, as we did above. Then move dⁿ to the left to get this.

xⁿ - dⁿ = ndy^n-1 + (n:2)d²y^n-2 + … + (n:2)d^n-2y² + nd^n-1y

Suppose n does not divide x y or z. Each binary coefficient is divisible by n, hence the right side is divisible by n, hence xⁿ = dⁿ mod n.

Raising to the n^th doesn't change a thing mod n, so x = d mod n.

consider the multiplicative group of units mod n². This is Z_n*Z_n-1. Raising x to the n^th power multiplies the log of x by n within this group. The first component drops to 0, and the second component is unchanged. Since x = d mod n, xⁿ = dⁿ mod n². The right side of our equation is divisible by n².

By assumption, n does not divide y. Since x is nonzero mod n, d is also nonzero mod n. In other words, n doesn't divide d either. Factor ndy out of the right side of the equation. What remains is divisible by n.

When n = 3 we are left with d+y. This is divisible by 3, yet z is not divisible by 3, so we have a contradiction. At least one variable in x³ + y³ = z³ is divisible by 3.

When n = 5 we find that 5 must divide y³ + 2dy² + 2yd² + d³.

Write this as y³+d³ + 2yd(d+y).

Since d+y = z, which is nonzero mod n, factor this out, leaving y² + dy + d². Divide through by y and set w = d/y to obtain 1 + w + w² = 0. Since this is impossible mod 5, 5 divides x y or z.

Let's try this trick with n = 7.

y⁵ + d⁵ + 3dy(y³+d³) + 5d²y²(d+y)

Pull out y+d to get this.

y⁴ + 2dy³ + 3d²y² + 2d³y + d⁴

Set d = 2y, and indeed it comes out 0. This approach doesn't work for all n.