RMO 400 points?
Find the sum of all values of , such that there exists some positive integer triples such that and satisfy the equation above.
The answer is 505.
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1 solution
Solution. First note that x divides 2012 · 2 = 2^3 · 503. If 503 | x then the right-hand side of the equation is divisible by 503^3, and it follows that 503^2 | xyz + 2. This is false as 503 | x. Hence x = 2^m with m ∈ {0,1,2,3}. If m ≥ 2 then 2^6 | 2012(xyz + 2). However the highest powers of 2 dividing 2012 and xyz+2 = (2^m)(yz)+2 are 2^2 and 2^1 respectively. So x = 1 or x = 2, yielding the two equations
y^3 +z^3 =2012(yz+2), and y^3 +z^3 =503(yz+1).
In both cases the prime 503 = 3 · 167 + 2 divides y^3 + z^3. We claim that 503 | y + z. This is clear if 503 | y, so let 503 ∤ y and 503 ∤ z. Then y^502 ≡ z^502 (mod 503) by Fermat’s little theorem. On the other hand y^3 ≡ −z^3 (mod 503) implies y^3·167 ≡ −z^3·167 (mod 503), i. e. y^501 ≡ −z6501 (mod 503). It follows that y ≡ −z (mod 503) as claimed. Thereforey+z=503k with k≥1. In view of y^3+z^3 =(y+z)((y−z)^2+yz) the two equations take the form k(y−z)^2 +(k−4)yz=8, (1) k(y−z)^2 +(k−1)yz=1. (2) In (1) we have (k−4)yz≤8, which implies k≤4. Indeed if k>4 then 1≤(k−4)yz≤8, so that y≤8 and z≤8. This is impossible as y+z=503k≥503. Note next that y^3+z^3 is even in the first equation. Hence y + z = 503k is even too, meaning that k is even. Thus k=2 or k=4. Clearly(1) has no integer solutions for k=4. If k=2 then(1) takes the form (y + z)^2 − 5yz = 4. Since y + z = 503k = 503 · 2, this leads to 5yz = 5032 · 22 − 4. However 5032 · 22 − 4 is not a multiple of 5. Therefore (1) has no integer solutions. Equation(2) implies 0≤(k−1)yz≤1, so that k=1 or k=2. Also 0≤k(y−z)2 ≤1, hence k=2 only if y=z. However then y=z=1,which is false in view of y+z≥503. Therefore k=1 and (2) takes the form (y−z)^2 =1, yielding z−y=|y−z|=1. Combined with k = 1 and y + z = 503k, this leads to y = 251, z = 252. In summary the triple (2, 251, 252) is the only solution.