Thread: solve the equation....

Find the integer solution for the following equation....

y power 2 = x power 3 − 432

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suresh

y = 36 and x = 12

hi mdk69,

Your answer is absolutely correct...Plz explain your solution.It will helps to others.

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suresh

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.

then how you solved it ...

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 non-zero 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.


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suresh

thanx for the explanation

You forced me to recall those cryptic rules...