Maths Olympiad Prep

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Problem 1243

AIME late
Number theory Difficulty 5.3 Prove it Ukrainian National Mathematical Olympiad, 3rd Round · Ukraine

How many integer solutions does the equation have
x2+y2+z2−xy−yz−zx=x3+y3+z3+s, x^2 + y^2 + z^2 - xy - yz - zx = x^3 + y^3 + z^3 + s,
if
a)s=0; a) s = 0;
b)s=1? b) s = 1?

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

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Official solution

a) Take x=−yx = -y, we get
3x2+z2=z3. 3x^2 + z^2 = z^3.
Take x=tzx = tz, we get: 3z2t2+z2=z33z^2t^2 + z^2 = z^3, or 3t2+1=z3t^2 + 1 = z, therefore
z=3t2+1,x=t(3t2+1),y=−t(3t2+1) z = 3t^2 + 1, x = t(3t^2 + 1), y = -t(3t^2 + 1)

b) Note that t3≡t(mod3)t^3 \equiv t \pmod{3}. Suppose that there exists solution and denote and consider equation modulo 33. We have
a+1≡x+y+z+1≡x3+y3+z3+1≡x2+y2+z2−xy−yz−zx≡≡x2+y2+z2+2(xy+yz+zx)≡(x+y+z)2≡a2, \begin{aligned} a+1 &\equiv x+y+z+1 \equiv x^3+y^3+z^3+1 \equiv x^2+y^2+z^2-xy-yz-zx \equiv \\ &\equiv x^2+y^2+z^2+2(xy+yz+zx) \equiv (x+y+z)^2 \equiv a^2, \end{aligned}
so a2−a−1=3a^2 - a - 1 = 3 which is impossible.

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