0

For integers $a,b$ that are relatively prime to one another, we have $a \mid t$ and $b \mid t$. It should follow that $ab \mid t$, but I cannot manage to prove this. Here is as far as I was able to get.

If $a \mid t$ and $b \mid t$, there exist integers $\beta, \alpha$ such that $$a \beta = t, \; b\alpha = t.$$ Hence, $$(a\beta)(b\alpha) = (ab)(\beta \alpha )=t^2,$$ so $ab \mid t^2$.

Why, though, does $ab$ divide $t^2$? Further, how do I even know that $ab \leq t$?

Another idea was to use Bezouts identity and assert the existence of integers $x$ and $y$ such that $$ax + by = 1.$$ Multiply through by $\beta \alpha$ and reassociate: $$(a\beta)(x\alpha) = (b\alpha)(y\beta) = \alpha \beta,$$ but this doesn't appear to help me.

John P.
  • 2,136

1 Answers1

1

The Bezout idea is good, try multiplying $ax+by=1$ by $t$ and using you factorisations of $t$ with $a$ and $b$.

Dave
  • 13,568
  • So from $ax + by = 1$, I multiply through by $t$ to get $axt + byt = t$. Using the expressions for $t$, I get $(ax)(b\alpha) + (by)(a\beta) = t$. Reassociating gives $(ab)(x\alpha) + (ab)(y\beta) = t$. Since $ab$ divides each term on the left-hand side, it divides the right-hand side. Hence, $ab \mid t$. Is this correct? – John P. Jun 21 '20 at 01:39
  • Looks good to me. – Dave Jun 21 '20 at 01:41