Math Problem #2 (Number Theory)

The only answer is when x=1 y=1 and x=-1 y=1
If x=0 then y=(1/4)^(1/3)
If x is negative, its the same as being positive because its squared
If x=2 y=(13/4)^(1/3) ---- not integer
If x=3 y=7^(1/3) ---- not integer
Because y is cubed, it cannot have any other integer values that match the equation. You would need (3x^2+1)/4 to be either 0,1,8,27,64,125, etc. None of those values lead to an integer x value

It's easy to see that x must be odd.

Set x=2m+1 and simplify to get 3m^2+3m+1=y^3.

We recognize that 3m^2+3m+1=(m+1)^3-m^3, which is the tricky step.

The equation is equivalent to y^3+m^3=(m+1)^3.

By the special case of Fermat's Last Theorem when the exponent is 3, we see that m=0 and y=1 or m=-1 and y=1.

Thus the only solutions to the original question are x=1,y=1 and x=-1,y=1

I guess it's sort of interesting that the solution involve's Fermat's last theorem. So before they started proving the special cases with computers, we, technically, wouldn't have a proof. Or is there another solution that doesn't involve FLT?

The special case of FLT with exponent 3 is quite easy compared to the general result: It was first proven by Euler centuries ago

The only solution is (-1,1) and (1,1).

Easy to get x is odd,so let x=2z+1 and do some algebra: y^3=3z^2+3z+1=(z+1)^3-z^3.

Look at the above Eq. again and recall FLT, the only solution is y=0//impossible or z=0//(1,1) solution or (z+1)=0//(-1,1) solution.

Hope it is clear.

if x is even, then 3x^2+1 will be odd and 4y^3 is always even.