A number theory problem by Eamon Gupta

$(a^2+b)-(a+b^2)=(a-b)(a+b-1)=\mathfrak Z$

For $\mathfrak Z$ to be a prime, one out of $a-b$ or $a+b-1$ must be $1$ . Setting $a+b-1=1$ gives $a=b=1$ and hence $\mathfrak Z=0$ which is not a prime , hence rejected. Now let's take $a-b=1$ or $a=b+1$ while $a+b-1=(b+1)+b-1=2b$ and its obvious that $2b$ is a prime only when $b=1$ and hence $a=b+1=2$ while $\mathfrak Z=(1)(2)=2$ .

$\large \therefore (1+2)^2=\boxed{\color{#007fff}{9}}$

WLOG $a^2+b > a+b^2$ :

$(a^2+b)-(a+b^2) = a^2-b^2-(a-b)=(a-b)(a+b-1) \Rightarrow \text{prime}$

All primes have 2 factors: 1 and itself, therefore one of the factors must be 1:

$a-b<a+b-1$ therefore $a-b = 1 \rightarrow a= b+1$

Substituting this: $((b+1)-b)((b+1)+b-1) = (1)(2b)=2b \rightarrow prime$

The only multiple of 2 which is prime is 2 itself; therefore $\boxed{b=1}$

$a = b + 1 \rightarrow \boxed{a=2}$

Therefore $(a+b)^2=\boxed{9}$