1. ## prove

prove that the system

(x^6)+(x^3)+(x^3)y)+y= (147^157)

(x^3)+((x^3)y)+(y^2)+y+(z^9)=(157^147)

has no solutions in integers x, y and z.

2. ## Brute Force

$x^6+x^3+x^3y+y= 147^{157}$
$x^3+x^3y+y^2+y+z^9=157^{147}$

The first equation $(x^3)^2+(y+1)x^3+(y-147^{157})=0$ is quadratic, so $x^3=\frac12(-(y+1)\pm\sqrt{(y+1)^2+4y-4*147^{157}})$

So $x^3$ can be substituted in the second equation $x^3(y+1)+y^2+y+z^9=157^{147}$. After simplifying, we arrive at...

$4\left(z^9+\frac12(y^2-1-2*157^{147})\right)^2=8y^3+(17-4*147^{157}+2*157^{147})y^2+$ $(6-8*147^{157})y-2(2*147^{157}+2*(157^{147})^2+3*157^{147}-1)$

(Someone may want to check my algebra for small errors.)

Since the LHS is a perfect square, it would be sufficient to show that no integer values of y allow the RHS to be a perfect square. One approach may be to attempt to factor the RHS as a perfect square minus a small constant, and then show that there are only a finite number of squares who differ by that constant -- the smaller the better. This may be implausible by hand.

Hope this brute force progress helps!