help

so we get y=4 and  x=3since the only integer for x^2+y^2=25 is 3,4,5 since $\sqrt{25}$=5 and by the pythagorean theorem either x or y=3 or 4 but to be specific we must check and ony 4=3^2-5 works so x=3 , y=4

We can actually solve the equation to see how many real solutions there are.

$\begin{cases}y = x^2 - 5\\x^2 + y^2 = 25\end{cases}\\ x^2 + (x^2 - 5)^2 = 25\\ x^4 - 9x^2 = 0\\ x^2 = 0 \text{ or } x^2 = 9\\ x = -3, 0, 3$

Plugging in these values into the equation gives

$y = 4, -5, 4$

in this order.

Therefore there are 3 pairs of real solution, namely $(-3, 4), (0, -5), \text{ and }(3, 4)$