Fermat’s Last Theorem (Proof for n = 4)

There are no nonzero integer solutions to x⁴ + y⁴ = z⁴

001	Assume x4 + y4 = z4, where x, y and z are nonzero integers.
002	Letting w = z2, we have x4 + y4 = w2.
003	Dividing out any common factors, there exists an X, Y, and W such that X4 + Y4 = W2 in which X, Y, and W are relatively prime.
004	Then (X2)2 + (Y2)2 = W2
005	Then there exists a Pythagorean solution for (X2)2 + (Y2)2 = W2, which is X2 = p2 – q2, Y2 = 2pq, and W = p2 + q2, where p, q and X are relatively prime.
006	Then there also exists a second Pythagorean solution for X2 = p2 – q2, which is X = m2 – n2, p = m2 + n2, and q = 2mn, where m and n are relatively prime.
007	Assume p and m are not relatively prime:
008	Then p = fP and m = fM
009	And n is also divisible by f since n2 = p – m2 = fP – (fM)2 = fP – f2M2 = f(P – fM2)
010	Then n and m have a common factor f.
011	But this contradicts that m and n are relatively prime.
012	So p and m are relatively prime.
013	Assume p and n are not relatively prime:
014	Then p = fP and n = fN
015	And m is also divisible by f since m2 = p + n2 = fP + (fN)2 = fP + f2N2 = f(P + fN2)
016	Then n and m have a common factor f.
017	But this contradicts that m and n are relatively prime.
018	So p and n are relatively prime.
019	Therefore, since m and n are relatively prime, and p and m are relatively prime, and p and n are relatively prime, that means p, m, and n are relatively prime.
020	Now  Y2 = 4pmn, because Y2 = 2pq = 2p(2mn) = 4pmn.
021	Since Y2 = 4pmn and p, m, and n are relatively prime, p, m, and n are squares, so p = w’2, m = x’2, and n = y’2
022	Substituting into p = m2 + n2, we have w’2 = (x’2)2 + (y’2)2, or x’4 + y’4 = w’2.
023	Now x’ < w since x’ is a factor of m which is a factor of q which (because W = p2 + q2) is less than W which a factor of w
024	And y’ < w since y’ is a factor of n which is a factor of q which (because W = p2 + q2) is also less than W which is a factor of w
025	And w’ < w since w’ is a factor of p which (because W = p2 + q2) is also less than W which is a factor of w
026	So there exists integers x’, y’, and w’ such that x’4 + y’4 = w’2 and 1 ≤ x’ < w, 1 ≤ y’ < w, and 1 ≤ w’ < w.
027	Therefore, if x4 + y4 = w2, there must exist other positive integers x’, y’, w’ such that 1 ≤ x’ < w, 1 ≤ y’ < w, and 1 < w’ < w in which x’4 + y’4 = w’2.
028	But we can use the same argument to prove the existence of more positive integers x”, y”, w” such that 1 ≤ x” < w’, 1 ≤ y” < w’, and 1 < w” < w’ in which x”4 + y”4 = w”2, and another x’”, y’”, w’”, and so on, so we have a contradiction by the method of infinite descent.
029	Therefore, there are no nonzero integer solutions to x4 + y4 = z4.

Q.E.D.