A number theory problem by soumyaranjan ram

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^myz + 2 are 22 and 21 respectively. So x = 1 or x = 2, yielding the two equations y^3 + y^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 y3 ≡ −z^3

(mod 503) implies y^3 167 ≡ −z^3 167 (mod 503), i. e.

y^501 ≡ −z^501 (mod 503). It follows that y ≡ −z (mod 503) as claimed. Therefore y + 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 · 2 2 − 4. However 503^2 · 2^2 − 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. so x= 2

$x^3(y^3+z^3)$ = 2012(xyz+2).

Divide by $x^3$ on both sides

We get ,

$y^3+z^3$ = $\frac{2012yz}{x²}$ + $\frac{4024}{x³}$ ,

As clearly L.H.S is an integer ,

So , considering that if 2012yz/x² as an integer

Then 4024/x³ must be integer .

4024 = (8)(503).

So x must be 2.(minimum).

Then 2012yz/x² will be equal to 503yz which is also integer.

And L.H.S $y^3+z^3$ ≥ 3yz. So it can be 503yz.