Integer roots of $x^2+y^2 = 25$ and $x^3+y^4=145$

Question

I'm trying to find the solution of $x^2+y^2=25$ and $x^3+y^4=145$.

I tried doing substitution, which leads me to:

$y^2=25-x^2$ substituted to $x^3+y^4=145$

$x^3+(25-x^2)^2=145$

$x^3+x^4-50x^2+480=0$

But it led me nowhere and I'm stuck.

A very detailed step will be much appreciated. Thanks!

Answer 1

First you want $x^3+x^4-50x^2+480=0$ more conventionally written $x^4+x^3-50x^2+480=0$

Then you can apply the rational root theorem - any integer $x$ must divide $480$.

Then you can apply your knowledge of $x^2+y^2=25$ to reduce the number of integers you have to test.

Or alternatively you can spot a solution.

Answer 2

Go with this: there are only eight integer solutions to $u^3 + 145 = v^2,$

(u,v)

u^3 + 145 = v^2

E_+00145: r = 2   t = 1   #III =  1
          E(Q) = <(-4, 9)> x <(-1, 12)>
          R =   1.6626192366
           8 integral points
            1. (-4, 9) = 1 * (-4, 9)
            2. (-4, -9) = -(-4, 9)
            3. (54, 397) = 1 * (-4, 9) - 1 * (-1, 12)
            4. (54, -397) = -(54, 397)
            5. (-1, 12) = 1 * (-1, 12)
            6. (-1, -12) = -(-1, 12)
            7. (6, 19) = -1 * (-4, 9) - 1 * (-1, 12)
            8. (6, -19) = -(6, 19)

see http://en.wikipedia.org/wiki/Mordell_curve and http://tnt.math.se.tmu.ac.jp/simath/MORDELL/ and http://tnt.math.se.tmu.ac.jp/simath/MORDELL/MORDELL+

Out of these, the only solution with $v$ a square is $(-4,9),$ so the only integer answers to $q^3 + 145 = r^4$ are $(-4,3)$ and $(-4,-3).$ Write $x=-q, y=r,$ you get $x^3 + y^4 = 145$ only for $(x=4, y=3)$ and $(x=4, y=-3).$

In both cases, $x^2 + y^2 = 25.$