Given two distinct primes, $p_1,p_2$, is it true that there are no non-zero integers $k_1,k_2$,$|k_1| < p_2$, $|k_2|<p_1$ such as that:
If so, how to prove it?
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The statement is true.
Let us assume that $k_1,k_2$ are integers with $|k_1| < p_2$, $|k_2|<p_1$ and $k_1p_1=k_2p_2$. $k_1$ and $k_2$ are products of prime numers, say $$|k_1|=a_1\cdots a_n,\quad |k_2|=b_1\cdots b_m.$$
Then you have the equation
$$ a_1\cdots a_n\cdot p_1=b_1\cdots b_m\cdot p_2.$$
After eventually dividing all equal primes of the $a_i$'s and the $b_j$'s of this equation (here you need, that $p_1, p_2$ are distinct, so that you cannot divide all $a_i$'s and $b_j$'s!) you can assume, that there must be an $a_r$ with $a_r=p_2$ or a $b_s$ with $b_s=p_1$. But then for the first case (the second case analogously)
$$|k_1|= a_1\cdots a_n\geq a_r=p_2,$$ which is a contradiction.
At the suggestion of OP:
First of all, what we are asked to prove is false, since nothing in the statement of the question prevents us from taking $k_1=k_2=0$. So let's rule that out by adding the condition $k_i\ne0$ for $i=1,2$.
Now there's a theorem that says if $p$ is prime and $p\mid ab$ then $p\mid a$ or $p\mid b$. This theorem is generally encountered as a step along the way to the Unique Factorization Theorem.
Given that theorem, from $k_1p_1=k_2p_2$ we have $p_1\mid k_2p_2$ so $p_1\mid k_2$ or $p_1\mid p_2$. Now from $0\lt|k_2|\lt p_1$ we can rule out the first alternative, and from $p_i$ being distinct primes, we can rule out the second, and we're done.