
Contributing Member
solve the equation....
Find the integer solution for the following equation....
y power 2 = x power 3 − 432

suresh

Junior Member
Re: solve the equation....

Contributing Member
Re: solve the equation....
hi mdk69,
Your answer is absolutely correct...Plz explain your solution.It will helps to others.

suresh

Junior Member
Re: solve the equation....
I put the eq. in the folowing form
Y^2  x^3 = 432
So i divided by prime numbers 432 = 2^4*3^3, then the result must be multiple of two's and three's. Maybe there is a scientific method, I don't know it. I think it was lucky.

Re: solve the equation....
then how you solved it ...

Contributing Member
Re: solve the equation....
Here is the way of solution....
Note that x3 = y2 + 432 is a perfect cube 63(y2 + 432) = 216(y2 + 432) is a perfect cube. But 216(y2 + 432) = (y + 36)3  (y  36)3.
Hence (6x)3 + (y  36)3 = (y + 36)3. (1)
By Fermat's Last Theorem, an + bn = cn has no nonzero integer solutions for a, b and c, when n > 2. Here we need the result only for the case n = 3, which was first proved by Euler, with a gap filled by Legendre.
However, x > 0.
Hence (1) can hold only when y  36 = 0 or y + 36 = 0; that is, y = ±36, in which case 6x = 72.
Therefore the only solutions are x = 12, y = ±36.

suresh

Re: solve the equation....
thanx for the explanation
You forced me to recall those cryptic rules...
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