Prove or disprove that if $a\mid(sb + tc)$ for all $s,t$ elements of integers, then $a\mid b$ and $a\mid c$

Question

Prove or disprove that if $a\mid(sb + tc)$ for all $s,t$ elements of integers, then $a\mid b$ and $a\mid c$

My question is "for all". I'm clearly misunderstanding something, because my intuition is to say, well this is obviously not true for every integer $s,t$ and I can provide a counter-example. However, what I just suggested to do should be done if it said "there exists $s,t$ that are elements of integers".

can someone walk me through it?

Answer 1

Let $s = 1, t= 0 \to a\mid b$, and let $s = 0, t=1 \to a\mid c$.

Alternatively, if for all $s,t$ that $a\mid (sb+tc) \to a\mid \text{gcd(b,c)} \to a\mid b,a\mid c$.