Question 900033
<pre>Let the two positive integers be p and q, p > q

p² - q² = 376

(p-q)(p+q) = 376

So (p-q) and (p+q) are two integers, (p-q) < (p+q),
whose product is 376.  The only pairs of factors
with product 376 are

1,376
2,188
4,94
8,47

So we have 4 possible systems of equations:

{{{system(p-q=1,p+q=376)}}},{{{system(p-q=2,p+q=188)}}},{{{system(p-q=4,p+q=94)}}},{{{system(p-q=8,p+q=47)}}}

The solution to the first system is (p,q) = (188.5,187.5),
and those aren't even positive integers.  So the smallest
possible positive difference of p and q can't be 1.

The solution to the second system is (p,q) = (95,93), and
the positive difference of them is 2.  So that's the answer.
We don't need to solve the other two systems, since p and q
differ by more than 2 in those.

Answes:  It's the positive difference of 95-93, which is 2.

Edwin</pre>