I know that if $a\mid b$ and $a\mid c$, then $a\mid sb+tc$ for all $s,t$.
Is this line below true? $$a\mid c\land\forall s,t:a\mid sb+tc\implies a\mid b$$
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And your conditions for the second line are for all $s,t$ right? – Parcly Taxel Sep 24 '17 at 11:07
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@ Parcly yes It is – xuoimai Sep 24 '17 at 11:09
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Since $a\mid c$, $a\mid tc$ for all $t$; since $a\mid sb+tc$, we can subtract $tc$ and obtain $a\mid sb$. Since this is true for all $s$, it is true for $s=1$, whereupon we have $a\mid b$.
Parcly Taxel
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Why not simply consider the case $s = 1$? That'd get you there even faster. – John Hughes Sep 24 '17 at 11:14
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I suspect that your interpretation of your own question is not the right one. For if you're given that $a \mid sb + tc$ for all integer $s$ and $t$, then choosing $s = 1, t = 0$ shows that $a \mid b$, and the other hypothesis involving $c$ isn't even needed.
I suspect that the right interpretation is this: suppose you know that for some $s$ and $t$, both $$ a \mid c \\ a \mid sb + tc. $$
Can you conclude that $a | b$?
The answer is no. For instance, pick $a = c = s = 2, t = 0, b = 1$. Then you have that $a \mid c$ because $2 \mid 2$, and $a \mid sb + tc $ because $2 \mid 2 \cdot 1 + 0 \cdot 2 = 2$, but it's not the case that $a \mid b$ because $2$ does not divide $1$.
John Hughes
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