One Simple Trick to Solve This National Maths Olympiad Problem

[wesleydeng]

Either p | x or p | y. If p | x, x=kp => ky^3=kp+y => k | y, y=mk => p = m(m^2*k^3-1) => m=1 => p=k^3-1. So only k=2, p=7, y=2, x=14.
For p | y, y=kp, => x=kp/(k^3*p^2-1). Easy to see x<1, no sol.

[MizardXYT]

If x and y are allowed to take on negative values, you also have the solution (x, y, p) = (-2, 1, 2)

[mcwulf]

Another good problem. I needed a couple of nudges from the video to solve this one.

[johns.]

I tried this with xy^2/x+y = p and got two sets of solutions.

[batapanos]

That's Greece 2015! I also suggest you do Greece national olympiad 2010 problem 1 which is one of my favourite

[moonlightcocktail]

1/p = 1/y2 (y+x/yx)
if y does not divide y + x, some factor m (not 1) of y will not be cleared out in the RHS and we will get m^2 | p, contradiction.
so x = ky for positive k. If x = y, (x^3)/2 = p, but then we need 2 | x => 4 | p, contradiction.
ky^3/(k+ 1) = p
k + 1 must be a factor of y^3 since k and k + 1 share no factors. y^3= m(k + 1)
km = p implies k = 1 (not possible) or y^3 = k + 1, k = y^3 - 1
so let y be free, x = (y^3 - 1)y. then we get
p = y^3 - 1
this is prime only for y = 2.
the only solution is thus
(14, 2, 7)

[SuperYoonHo]

thanks so much
sorry i am late but this video is superb!
the reason is because we didn't have internet for 2 days