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If we know the value of $a\pmod {p_1}$ and $b\pmod {p_2}$ where $p_1$ and $p_2$ are prime, is it possible to find the value of $ab\pmod {p_1p_2}$ ? . I was thinking of using the chinese remainder theorem but I can't seem to generalize it.

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    If $p_{1} \ne p_{2}$, use CRT to get the values of $a$ and $b$ modulo $p_{1} p_{2}$, then you're done. If $p = p_{1} = p_{2}$, you cannot determine the result, as you can see by taking first $a = b = 1$, and then $a = 1 +p$, $b = 1$. – 2017-01-10
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    @AndreasCaranti sorry for the inconvenience but I have changed the question a bit. – 2017-01-10
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    No problemo ;-) – 2017-01-10

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Since you mention CRT, I assume $p_{1} \ne p_{2}$.

The answer is no.

Think of $a = 1 + s p_{1}$ (all values are congruent modulo $p_{1}$), $b = 1 + t p_{2}$ (all values are congruent modulo $p_{2}$), which yield all the products $$ ab \equiv 1 + s p_{1} + t p_{2} \pmod{p_{1} p_{2}}. $$

For instance when $p_{1} = 3$ and $p_{2} = 5$ we have $$ 1 \cdot 1 = 1 \not\equiv 9 \equiv (1 + 3) (1 + 5) \pmod{15}. $$

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    Note that since $\,p_1,p_2\,$ are coprime, by Bezout we can choose $\,s,t\,$ so $\,sp_1+tp_2\,$ is any integer, therefore $\,ab\,$ can take *every* value mod $\,p_1p_2.\ $ – 2017-01-10
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    @BillDubuque, great comment, thanks! – 2017-01-10