Prove that for any natural $a$ there is a natural $b$ so that $a$ and $b$ are coprime and $a+b^2$ is not a prime

Question

Is there a simple proof of the following statement?

For any $a \in N$ exists such $b \in N$ that $a \perp b$ and $a + b^2$ is a composite number.

Answer 1

Pick $b=2a+1$: $$ a+(2a+1)^2 = (a+1)(4a+1). $$

Answer 2

For $a>1$ odd we can take $b=1$. Then $a+b^2=a+1>2$ is even, hence not prime. In general, following Jack, one could also take $b=6a+1$. Then $$ a+b^2=(9a+1)(4a+1). $$

Answer 3

Any $b$ congruent to $1$ modulo $a+1$ will do the job. Such a $b$ will be of the form $k(a+1)+1=ka+(k+1)$, so any $k$ such that $k+1$ is coprime to $a$ will work, for example $k=a$.

In fact, for $k=a$ we get $b=a(a+1)+1$ and $$a+b^2=a+a^2(a+1)^2+2(a+1)+1=(a+1)+a^2(a+1)^2+2(a+1)=(a+1)(a^3+a^2+2a+1)$$

which is obviously divisible by $(a+1)$.