*Post by Jerry Kraus**Post by bassam king karzeddin*x^{2n} + y^{2m} = z^2

Doesn't have any integer solution

http://mathrefresher.blogspot.com/2005/05/coprime-numbers-xn-yn-zn.html

I am beginning to wonder a bit if the conjecture of the original post is in fact accurate, or provable.

I do not understand how what is in that post has anything to do with anything. The argument just shows that *IF* there is a solution with x, y, and z nonzero integers, THEN there is a solution with x, y, and z coprime. The argument to establish that can be done much more straightforwardly than the way given there:

Suppose p is a prime that divides at least two of x, y, and z. We claim that p divides the third.

If p divides x and y, then it divides both x^n and y^n, hence it divides x^n+y^n = z^n. Since p divides z^n, and p is prime, then it divides z. Thus, p divides all three.

If p divides x and z, then p divides x^n and z^n, hence it divides z^n - (x^n) = y^n. Since p is prime, then it divides y. Thus, p divides all three.

If p divides y and z, then p divides y^n and z^n, hence it divides z^n - (y^n) = x^n. Since p is prime, it then divides x. Thus, p divides all three.

Hence, if a prime p divides at least two of x, y, and z, then we can write x=pa, y=pb, z=pc, and we have

p^na^n + p^nb^n = p^nc^n; hence p^n(a^n+b^n) = p^nc^n.

Cancelling p^n, we get a^n + b^n = c^n, so (a,b,c) is also a solution.

Repeating this process as often as necessary, it follows that we may replace any solution with a solution (a,b,c) in which for every prime p, at MOST one of a, b, and c is divisible by p. Therefore, IF a solution exists, then there is a solution in which a, b, and c are pairwise coprime.

How does this relate to the situation at hand?

It is not clear that IF a solution exists then there must be a solution in which all three are coprime, because of the difference in exponents. If you try to argue as before, you would get:

Suppose (x,y,z) is a solution to x^{2n} + y^{2m} = z^2; if p is a prime, and p divides at least two of x, y, and z, then it divides at least two of (x^n, y^n, z), which is a solution to a^2 + b^2 = c^2, and hence it must divide all three. We can write x= pa, y=pb, z=pc. Then we get plugging in we get

p^{2n}a^{2n} + p^{2m}b^{2m} = p^2c^2.

Cancelling p^2 (which we can do because we are assuming m,n are at least positive) we get

p^{2(n-1)}a^{2n} + p^(2(m-1))b^{2m} = c^2.

If n,m>1, then This implies that p divides c^2, hence c; so we can write c=pd, and get

p^{2(n-1)}a^{2n} + p^{2(m-1)}b^{2m} = p^2d^2

and cancelling we can reduce by a further 2 to get

p^{2(n-2)}a^{2n} + p^{2(m-2)}b^{2m} = d^2.

We can continue doing this for min(n,m) steps. If n=m, then we will obtain

a^{2n} + b^{2m} = r^2

for some r, and we have obtained a smaller solution that has one fewer prime factors for a and b.

If n=/=m, then we will exhaust one of the powers but not the other, and the resulting equation will not necessarily yield a solution to the original equation.

Thus, I think that the best we can say in this situation is that in an arbitrary solution and for a given prime p, either all of x, y, and z are divisible by p, or else at most one is; that is, it cannot be that exactly two of them are divisible by p.

That does not preclude asking about the case in which for every prime we fall in the second case: that is, a solution with x, y, and z coprime (which would make x^n, y^m, and z a primitive pythagorean triple).

--

Arturo Magidin