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Number theory Difficulty 6.0 National olympiad Prove it

20. Prove: For any nNn \in \mathbf{N}^{*}, the indeterminate equation x2+y2=znx^{2}+y^{2}=z^{n} has infinitely many positive integer solutions (x,y,z)(x, y, z).

Solution

20. An interesting construction is as follows: x+yi=(a+bi)nx+y \mathrm{i}=(a+b \mathrm{i})^{n}, where i is the imaginary unit. When nn is fixed, there are infinitely many pairs of integers (a,b)(a, b) such that integers x,yx, y satisfy xy0x y \neq 0. In this case, x2+y2=(a2+b2)nx^{2}+y^{2}=\left(a^{2}+b^{2}\right)^{n} (this can be obtained by taking the modulus of both sides).

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.