Given $2$ integers $a,b$ which sum to a prime, is it possible to find a new integer $x$, so that $a+ bx$ is NOT prime?

Question

We start with $2$ unknown integers $a,b$.
It is known that the sum of $a$ and $b$ is prime and $a$ is not negativ and $b$ is greater than $0$.
$a+b$ is prime, where $a \ge 0, b > 0$
Is it possible to find a non negativ integer $x$ depending on $a$ and $b$, so that $a + x\cdot b$ is NOT prime ?

Answer 1

Given any positive integer $k,$ if $x=1+ (a+b)k,$ then $$\begin{align}a+bx&=a+b(1+(a+b)k)\\&=a+b+b(a+b)k\\&=(a+b)(1+bk).\end{align}$$

This gives infinitely many $x$ so that $a+bx$ is not prime and, specifically, divisible by $a+b.$

This can be written in terms of congruence:

$$x\equiv 1\pmod {a+b}\implies \\a+bx\equiv 0\pmod {a+b}$$

In particular, when $x>1,$ $a+bx>a+b$ and is divisible by $a+b.$

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More generally, if $p(x)$ is a non-constant integer polynomial with non-negative coefficients, and $p(1)$ is prime, then for $k>0,$ $p(1+kp(1))$ is always divisible by $p(1)$ and greater than $p(1),$ so $p(1+kp(1))$ is composite.

Answer 2

Polynomial remainder theorem is the name of the theorem we use. It's basic use here, is that it tells us if we shift a variable in a polynomial by a set value, the remainder on division by that value will be the same as the remainder prior to the shift. That means any polynomial you have $f(n)$ , has $f(h)\mid f(h)\implies f(h)\mid f(f(h)+h)$