IMO 2012 Shortlist N2

Category: Number Theory

Problem

Find all triples (x,y,z) of positive integers such that x ≤ y ≤ z and x3 (y3

z3 ) = 2012(xyz + 2). Solution. First note that x divides 2012·2 = 23 ·503. If 503 | x then the right-hand side of the equation is divisible by 5033 , and it follows that 5032 | xyz + 2. This is false as 503 | x. Hence x = 2m with m ∈ {0,1,2,3}. If m ≥ 2 then 26 | 2012(xyz + 2). However the highest powers of 2 dividing 2012 and xyz + 2 = 2m yz + 2 are 22 and 21 respectively. So x = 1 or x = 2, yielding the two equations y3
z3 = 2012(yz + 2), and y3
z3 = 503(yz + 1). In both cases the prime 503 = 3 · 167 + 2 divides y3
z3 . We claim that 503 | y + z. This is clear if 503 | y, so let 503 ∤ y and 503 ∤ z. Then y502 ≡ z502 (mod 503) by Fermat’s little theorem. On the other hand y3 ≡ −z3 (mod 503) implies y3·167 ≡ −z3·167 (mod 503), i. e. y501 ≡ −z501 (mod 503). It follows that y ≡ −z (mod 503) as claimed. Therefore y + z = 503k with k ≥ 1. In view of y3
z3 = (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 y3
z3 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 · 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.44