Complex Extensions, The Coprime Lemma

The Coprime Lemma

If any two terms in xⁿ + yⁿ = zⁿ have a common prime factor p, p can be divided out of all three terms. If there is a solution, there is a minimal solution with coprime variables. This was described in the previous section.

We are going to generalize this result here. Consider the following equation.

ax^l + by^m = czⁿ

If c includes the factor pⁿ, we can multiply z by p and divide c by pⁿ. Thus it is safe to assume the primes in any coefficient are raised to powers smaller than the associated exponent.

In what follows, we will assume that the primes in any coefficient are raised to powers smaller than all three exponents. Also, let two of the exponents be equal, while the third is a factor of the other two. Consider the following example.

7x⁶ + 12y⁶ = 35z³

If p divides y and z, the coefficient on x cannot absorb all these powers of p, hence p divides x. If p divides two variables it divides the third, so assume p divides all three variables.

The left side is divisible by p⁶. If z is not divisible by p² then the right side cannot be divisible by p⁶, which is a contradiction. Therefore, divide x and y by p, and z by p², giving a smaller solution. This continues until the variables are pairwise coprime.

Coprime Coefficients

If the coefficients are pairwise coprime, which is not the case in the example shown above, each coefficient is coprime to the other two variables. Here is a new example.

7x⁸ + 15y⁸ = 11z²

If y is divisible by 7 then so is z, and we have reproduced the earlier case. Divide x and y by 7, and z by 7⁴, to find a smaller solution. In the smallest solution, variables are coprime to each other, and each variable is coprime to the other two coefficients.

Shared Powers of p

Assume the coefficients are all set to 1, and the exponents are unconstrained. How many powers of p are common to more than one variable? Consider the following example.

x³ + y³ = z⁴

In the following, x y and z are variables, while expressions like x³ are terms.

If the terms, or equivalently the variables, are coprime, there is nothing to talk about. So assume two terms are divisible by p, whence the third term is divisible by p as well. Can this be a minimal solution?

Let p^j divide each term, where j is maximal. If the first term is divisible by p^j, and the second by p^j+1, and the third by p^j+7, divide through by p^j and look mod p. The first term is nonzero and the other two are zero. This is impossible. Therefore, at least two of the three terms are divisible by precisely p^j. The third supports at least j instances of p.

If p^j divides all three terms, and there are no factors of p beyond j, then the exponents all divide j. In our example, j might be 12. Divide x and y by p⁴, and z by p³. This is a smaller solution. When the solution is minimal, exactly one of the three terms contains more than j instances of p.

If two terms, with exponents l and m, have j factors of p, then j is a multiple of l and m. Let j = lcm(l,m) for convenience. If the third exponent n divides into j, divide x by p^l/j, y by p^m/j, and z by p^n/j, and the equation still holds, producing a smaller solution.

This is often used when two of the three exponents are equal, as shown in our example. Suppose the second and third terms are divisible by precisely p^j, while the first term is divisible by more than p^j. Now j is a multiple of 3 and 4, so call it 12. The first exponent is 3, just like the second. It is going to divide the lcm of 3 and anything; in this case it divides into 12. So we can divide the variables by various powers of p and find a smaller solution. In a minimal solution, the first two terms, carrying the same exponent, have the same powers of p. This is illustrated by the minimal solution 9³ + 18³ = 9⁴. Since x and y share a common factor of 3, they are both divisible by the same power of 3, namely 9.

Here is another example of a minimal triple whose variables are all divisible by 3. In this case the first two variables have only one factor of 3 each.

33³ - 6³ = 189²