Irrational set equations

Assume that $y=0$ . The first equation with $y=0$ implies that $x=0$ , but $x=y=0$ does not satisfy the second equation, so $y\ne 0$ .

Let $x=ky$ . From the first equation $\left(x^2+y^2\right)=y^4\left(y^2+1\right)$ , replace $x$ with $ky$ will change it to $ky\left(k^2y^2+y^2\right)=y^4\left(y^2+1\right)$ or $k^3y^3+ky^3=y^6+y^4$ , which leads to $y^3\left(k^3+k\right)=y^3\left(y^3+y\right)$ . We have proven that $y\ne 0$ above, so $k^3+k=y^3+y$ .

Next, we will compare $y$ with $k$ :

*If $k<y$ then $k^3<y^3$ , implying $k^3+k<y^3+y$ , which is not correct.

*If $k>y$ then $k^3>y^3$ , implying $k^3+k>y^3+y$ , which is not correct.

*If $k=y$ then $k^3=y^3$ , implying $k^3+k=y^3+y$ , this matches what we have proven above.

Because of this, $k=y$ , but we let $x=ky$ at the beginning, so $x=y^2$ .

From the second equation $\sqrt{4x^2+5}+\sqrt{y^2+3}=5$ , replace $x$ with $y^2$ will change it to $\sqrt{4y^4+5}+\sqrt{y^2+3}=5$

We know that both $y^4$ and $y^2$ are always non-negative for all numbers $y$ , so this time we will compare $y^2$ with 1:

*If $y^2<1$ then $y^4<1$ , so $4y^4+5<9$ and $y^2+3<4$ , implying $\sqrt{4y^4+5}+\sqrt{y^2+3}<5$ , which is not correct.

*If $y^2>1$ then $y^4>1$ , so $4y^4+5>9$ and $y^2+3>4$ , implying $\sqrt{4y^4+5}+\sqrt{y^2+3}>5$ , which is not correct.

*If $y^2=1$ then $y^4=1$ , so $4y^4+5=9$ and $y^2+3=4$ , implying $\sqrt{4y^4+5}+\sqrt{y^2+3}=5$ , this is correct.

So $y^2=1$ , we know that $x=y^2=1$ as proven above, and from $y^2=1$ , we also know that $y=1$ because y is not a negative number.

Final result: $x+y=1+1=2$ .