RMO 2015 8

We first of all sum up all $(x + j)^{3} and (x + 7 - j)^{3}$ to form total 4 pairs by using the identity $a^{3}+ b^{3}$ we get the equation as ; $(2x + 7)(4(x)^{2} + 28x + 112)=(y)^{3}$ now we further factorize it to $(2x + 7)[(2x + 7)^{2}+ 63] = y^{3};; 2x + 7 = k$ then we have ; $(k^{2} + 63)k = y^{3}$ from here we can say y > k also now if we suppose k | y then $y^{3} - k^{3} = 63k$ $k^{3} | y^{3} - k^{3}$ so $k^{2} | 63$ so we can have k = -1,-3,+1,+3 and x = -4,-3,-5,-2 by putting all values of k we get corresponding values of y = -4,-6,+4,+6. Hence the sum is -14. If k does not divide y then this would mean that there exists a prime p such that $E_{p} k = a;; E_{p} y = b$ and a > b a < 3b then this would mean that $p | k^{2} + 63,, p|k,, p| 63$ . So now if p = 7 or 3 let $E_{7} y = a ::: E_{7} k = 3a - 1$ also if $E_{3}y = b ::: E_{3}k = 3b - 2$ . If suppose k has any other prime factors then all of them must have powers equal to their powers in $y^{3}$ so $k(k^{2} + 63)) = y^{3}$ now after cancelling k * 63 from the equation we have $[7^{3(2a - 1)+ 1}. 3^{3(2b-2) +2}. P^{6d}..........r^{6j} + 1] = c^{3}$ where p......r are other primes dividing k. So notice that $c^{3} - 1 = 63t^{3}$ where $t = 7^{2a -1}.3^{b - 2}.p^{2d}........r^{2j}$ this equation has no solution except c = 4 and t = 1 and hence for the main equation we have no solution if k does not divide y.

We have: x+7 > x+6 > ... > x

=> 8 (x+7)^3 > y^3 > 8 (x^3)

=> 2(x+7) > y > 2x

=> y belongs to {2x+1, 2x+2, 2x+3, ..., 2x+13} (since x and y are both integers)

We have y^3 = 8 (x^3) + 84 (x^2) + 420x + 784

In each case when we replace y with 2x+1, 2x+2, ..., 2x+13, all we have left is a quadratic function, which is easy to find if its roots are integers or not.

After taking all the cases, we have the following (x,y): (-2,6) , (-3,4) , (-4,-4) , (-5,-6)

Hence, the sum of all integers solution of x,y satisfying the equation above is -14

(-3,4),(-4,-4),(-2,6),(-5,6) for ordered pairs (x,y) .