A Diophantine Equation with Infinitely Many Solutions

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it is good to see how you confidently tackle and solve such a difficult problem, excellent job Syber, bring it on, I am here to learn bro

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Because the least common multiple of the exponents 2, 3, and 4 is 12, let x = 3^(6n), y = 3^(4n), z = 3^(3n), and t = 3^s (I have obtained the first three exponents by dividing 12n by the given power on x, y, and z, respectively). Note that the linear equation 12n + 1 = 11s has infinitely many solutions (n, s), such that n and s are natural numbers due, mainly, to the fact that gcd(11, 12) divides the constant term, which is one in this case. I claim that (x, y, z, t) = (3^(6n), 3^(4n), 3^(3n), 3^s) is a (natural number) solution to x^2 + y^3 + z^4 = t^11 for each ordered pair of natural numbers (n, s) that satisfies 12n + 1 = 11s. To verify this claim, note that, in this case, x^2 + y^3 + z^4 = 3^(12n) + 3^(12n) + 3^(12n) = 3^(12n + 1) = 3^(11s) = t^11.

richardryan