Find the integers solutions to:
2x² -3x=3y²
Thank you.

Silvia Ramos
IV Olimpiada Iberoamericana.

Hi,

I think that if x is positive then there are no solutions. 2x² - 3x = 3y² imples x,y must both be multiples of 3. So set x = 3l, y = 3m and you get l(2l - 1) = 3m² . Now l and 2l - 1 are coprime so it follows that either:

i) l = 3r² , 2l - 1 = s²
or
ii) l = r² , 2l - 1 = 3s²

for some integers r,s. To see this, think about the prime factorisation of l and 2l - 1, and use the fact their product is three times a square.

Now i) implies that 6r² - 1 = s² which is impossible since there is no natural s satisfying s² = -1 (mod 3). ii) implies that 2r² - 1 = 3s² , or r² = -1 + 3(2s² - r² + 1), which is again impossible since we cannot have r² = -1 (mod 3). Hence there are no solutions for x positive.

Clearly x = 0 gives the solution (0,0). I'm not totally certain at the moment about negative x.

The only solutions (x,y) where -1000000 < = x < 0 are:

(-3,±3)
(-36,±30)
(-363,±297)
(-3600,±2940)
(-35643,±29103)
(-352836,±288090)

I suspect there are infinitely many solutions, but I'll leave someone else to fill in the details... :-)

James.

PS. Do you reckon it's a coincidence that all the x-values are approximately 3.5 times 10ⁿ for some n?

This equation can be solved through Pell's equation. First, we have x=-3l, y=3m, some l,m positive integer. Putting it through, we get

2l² +l=3m²

Multiply both sides by 8 and add 1, we get

(4l+1)² -24m² =1

and now we quote that a Pell's equation having a non-trivial solution has infinitely many of them. There is a recurrence relation that will give rise to this. However, it is easy to work out the continued fraction of sqrt(24) and seeing that each convergent gives a solution to the equation A² -24B² =±1.

Kerwin

Nice trick there Kerwin....

love arun