KVPY 2016 SA Question 19

$n=10a+b=a^2+b^3$ where $0 < a \leq 9$ and $0 < b \leq 9$

$a(10-a) = (b-1)(b)(b+1)$

Clearly, $24 | (b-1)(b)(b+1)$

So, $24 | a(10-a)$

$3 | a(10-a) \Rightarrow a$ can take values $1,3,4,6,7,9$

$8 | a(10-a) \Rightarrow a$ can take values $2,4,6,8$

So, $24 | a(10-a) \Rightarrow a$ can take values $4,6$

The only solutions are $(a,b)=(4,3),(6,3)$

$43=10\times4+3=4^2+3^3$

$63=10\times6+3=6^2+3^3$

For $0 < a \leq 9$ and $0 < b \leq 9$ ,

$\begin{aligned} n=10a+b&=a^{2}+b^{3} \\ 0&= a^{2}-10a+b^{3}-b \end{aligned}$

$\begin{aligned} D &\geq 0 \\ (-10)^{2}-4.1. (b^{3}-b) &\geq 0 \\ 100 &\geq 4.1. (b^{3}-b) \\ 25 &\geq b^{3}-b \end{aligned}$

$\begin{aligned} n=10a+b&=a^{2}+b^{3} \\ 10a-a^{2} &=b^{3}-b \\ (10-a)a&=b^{3}-b \implies b^{3}-b \geq 9 \end{aligned}$

$\implies 9 \leq b^{3}-b \leq 25$

The only solution for $b$ is $3$ with $a=4$ or $a=6$

So, there are $\boxed{2}$ numbers, $43$ and $63$ .