Maths Olympiad Prep

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Algebra Difficulty 5.7 AIME, harder Prove it

5. Prove: The indeterminate equation
(7a+1)x3+(7b+2)y3+(7c+4)z3+(7d+1)xyz=0(7 a+1) x^{3}+(7 b+2) y^{3}+(7 c+4) z^{3}+(7 d+1) x y z=0

has only the trivial solution x=y=z=0x=y=z=0.

Solution

5. Let's assume (x,y,z)=1(x, y, z)=1, and note that u30,±1(mod7)u^{3} \equiv 0, \pm 1(\bmod 7). If there are non-trivial solutions, consider the congruence equation modulo 7, and discuss the cases (z,7)=1,(z,7)=7(z, 7)=1,(z, 7)=7 separately.

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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.