Arithmetic Partial But General Case Of Fermat's Last Theorem For n \gt 2 odd, the equation x^{n} + y^{n} = z^{n} has no solution with x+y prime. Proof → (This problem has been included in the XXVI Moscow Mathematical Olympiad (1963), final stage, junior grade.)





































































Proof Assume there is a solution. Since n is odd, x^{n} + y^{n} = (x+y)(x^{n-1} - x^{n-2}y + x^{n-3}y^{2} - \ldots - xy^{n-2} + y^{n-1}), which implies, in particular, that x+y divides x^{n} + y^{n} = z^{n}. But, by the assumption of the problem, x+y is prime and, since it divides z^{n}, it also divides z itself. But then (x+y)^{n} divides z^{n} = x^{n} + y^{n}. This is a contradiction because necessarily (x+y)^{n} \gt x^{n} + y^{n}.











































