Rational Gaze

Given: $x^y=y^3+10x+x^2$
$\implies x^y-10x-x^2=y^3$
$\implies x|y$
Let $y=kx$
Putting this in our equation, we get,
$x^{kx}-10x-x^2=k^3 x^3$
$\implies x^{kx-3}-\frac{(10x+x^2)}{x^3}=k^3$ or $y=1,2$ or $x=0$
Checking $y=1,2$ we don't get natural $x$ , so we reject $y=1,2$
Also, for $x=0$ , we get $y=0$ , but putting both of them together in the original equation will yield $0^0$ in L.H.S. , which is undefined, so we reject $x=0$
$\implies x^{kx-3}-\frac{(10x+x^2)}{x^3}=k^3$
$\implies x^{kx-3}-\frac{(10+x)}{x^2}=k^3$ (Cancelling $x$ from Numerator and Denominator)
$\implies x^2|(10+x)$
$\implies x^2 \le (x+10)$
For natural $x$ , we get $x=1$ or $x=2$ as the only numbers satisfying the restrictions.
If $x=1$ ,
$\implies y^3=-10$
Thus rejecting $x=1$ , we get $x=2$ and
$2^y=y^3+24$
We can easily plot the graph of L.H.S. and R.H.S. on the same cartesian plane and see that there are exactly two points of intersection out of which one gives negative value of $y$ and the other one is $y=10$ (To do this easily some calculus can be used).
Thus $\beta = 10$ , $\alpha = 2$
$\implies \frac{\beta}{\alpha}=\boxed 5$ .

Alternate solution
Like above, we find $x=2$ and get
$2^y=y^3+24$
Now,
Since L.H.S. of the equation is even $\implies$ R.H.S. is even
$\implies y$ is even.
Let $y=2k$
$\implies 2^{2k}=8k^3+24$
$\implies 2^{2k-3}=k^3+3$ (Since $y \ge 3$ from first part i.e. of finding the value of $x$ )
Since L.H.S. of the equation is even $\implies$ R.H.S. is even
$\implies k$ is odd.
Again rewriting the equation, we get,
$4^k=8k^3+24$
Also, since
$4 \equiv (-1) \mod {5}$
$\implies 4^k \equiv (-1)^k \equiv (-1) \mod {5}$ (Since $k$ is odd)
$\implies 8k^3 +24 \equiv (-1) \mod {5}$
$\implies k \equiv 0 \mod {5}$
Let $k=5j$
$\implies (4^5)^j=1000j^3+24$
$\implies (1024)^j=1000j^3+24$
Now, it is trivial to see that $j=1$ is a solution giving $y=2$ and for higher values of $j$ we can see even without graph that L.H.S. increases much more rapidly than R.H.S. . Further, we do not need to analyse for $j<1$ since that would give non-natural $y$ .
Therefore, again we get
$\beta = 10$ , $\alpha = 2$
$\implies \frac{\beta}{\alpha}=\boxed 5$ .

We are given $x^y = y^3 + 10x + x^2$ .

If $y = 1$ , then clearly $LHS < RHS$ .
If $y = 2$ , then clearly $LHS < RHS$ .
Henceforth, $y \geq 3$ .

If $x \geq 4$ , then for $y \geq 3$ , we have $x^y > y^3 + 10 x + x^2$ , so no solutions.

If $x = 3$ , then for $y \geq 4$ , we have $3^y > y^3 + 39$ . For $y = 3$ , we have $27 < 27 + 39$ , so no solutions.

If $x= 2$ , then we have $2^y = y^3 + 24$ . For $y \geq 11$ , we have $2^y > y^3 + 24$ . Hence, we only need to check $y = 1$ to 10, of which $y = 10$ works.

If $x = 1$ , then we have $1 = y^3 + 10 + 1$ which has no solutions.

Hence, $(2,10)$ is the only solution.

Clearly, x is the smaller number. If x=1, $y^3$ =10, not an integer. x=2, we have, $f(x,y)=y^3+10x+x^2-x^y=0,~\therefore~f(2,y)=y^3+20+4-2^y=0,\\\implies~y^3=2^y-24.~y>1,~~\therefore~2^y>24,\\\implies~y>4.~~Let~g(y)=2^y-24. ~~\text { Since y is a +tive integer,}\\ \text {and answer box excepts only integers, we check for y=6,8,10.} \\g(10)=1024-24=1000=10^3.~~~~~~~~requred~ratio= \dfrac{10} 2 = ~~~\Large \color{#D61F06}{5}\\```\\$
We can try to get a rough estimate as follows. $X^Y>Y^3~ \therefore~ YlogX>3logY,~~\implies~\dfrac Y {logY}>\dfrac 3 {logX}. \\ \text {Draw two graphs, one for the left and other for the right.}\\\text {They intersect near integer x=3 ONLY. }\\ \text{Since it is inequality we go for x=2. Then go as above.}$