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