Show that the equation 1 / x4+ 1 / y 4 = 1 / z 2 has no solution in positive integers.
Answers
Step-by-step explanation:
We have to show that the Diophantine equation x^4 + y^4 = z^2 (1) does not have an integer solution. Let us assume now that (1) has an integer solution (x1, y1, z1) as its smallest positive integer solution. Since the positive integers of this solution are relatively prime in pairs, without loss of generality, we can assume that x1 is odd, y1 is even and z1 is odd.
If we will show that equation (1) does not have a positive integer solution, then it follows that the equation x^4 + y^4 = z^4 (2) does not have an integer solution and also the equation x^n + y^n = z^n (3) for n = 4k, will not have integer solution when k is a natural number such that k ≥ 1. This follows directly from the fact that if the equation x^(4k) + y^(4k) = z^(4k) => (x^k)^4 + (y^k)^4 = (z^k)^4 had an integer solution for any k ≥ 1, then the known Diophantine equation X^4 + Y^4 = Z^4 with X = x^k, Y = y^k, Z = z^k would also have a positive integer solution. This means that the numbers x^k, y^k, z^k would be all positive integers, so X, Y, Z will also be positive integers. But this is impossible because of the equation (2).
In order to show that (2) does not have an integer solution, it is sufficient to show that the Diophanttic equation x^4 + y^4 = z^2 does not have a positive integer solution.
We now assume that equation (1) has the solution (x1, y1, z1), so that (x1, y1) = 1, (y1, z1) = 1 and (x1, z1) = 1 and we consider that this solution is the smallest possible. Without damage of generality, we can assume that x1 is odd, y1 is even and consequently z1 is odd. If ((x1)^2)^2 + ((y1)^2)^2 = (z1)^2 is true, then the mathematical formula of the solution of this diophantine equation implies that there exist positive integers a, b with a > b and a - b ≠ 2m (m being a nonzero natural number), such that:
(x1)^2 = a^2 - b^2 (3)
(y1)^2 = 2ab (4)
(z1)^2 = a^2 + b^2 (5)
From relation (3) it follows that (x1)^2 + b^2 = a^2 (6) and therefore we observe that the integers x1, b and a form a pythagorean triple with x1 odd and b even. Therefore, there exist relatively prime positive integers r, s with r > s and r - s ≠ 2m’ where m’ is a nonzero natural number, such that:
x1 = r^2 - s^2 (7)
b = 2rs (8)
a = r^2 + s^2 (9)
Replacing equations (8) and (9) in equation (4), we obtain:
(y1)^2 = 2(r^2 + s^2) 2rs => ((y1)/2)^2 = (r^2 + s^2)rs (10)
However, because r, s are relatively prime, it follows that G.C.D. (r^2, s^2) = 1 and therefore the numbers (r^2 + s^2), r and s will also be pairwise relatively prime. Therefore, we will have (r, s) = 1, ((r^2 + s^2), r) = 1 and ((r^2 + s^2), s) = 1 and from the relation (10), it follows that r, s and (r^2 + s^2) will all be perfect squares. This means that there exist positive integers t, u, q such that:
r = t^2 (11)
s = u^2 (12)
r^2 + s^2 = q^2 (13), therefore:
q^2 = r^2 + s^2 = c^4 + d^4 => c^4 + d^4 = q^2 (14)
Since we have 0 < q ≤ a < a^2 + b^2 = z1, there is another positive integer solution for the equation (1), consisting of the triple of positive integers c, d and q, which are pairwise relatively prime and smaller than x1, y1, z1. However, this cannot be true, because we have accepted that the solution consisting of the numbers x1, y1, z1 is the smallest possible. So the equation (1) does not have an integer solution and therefore the equation (4) does not have an integer solution too.