Dutch Mathematical Olympiad

a. Suppose the two positive integers are $n-10$ and $n+10$, and hence $n>10$. Then the product is equal to $(n-10)(n+10)=n^2-100$ and we are looking for an $n>10$ such that $n^2-100+23$ ends in the digits $23$. But that means we want $n^2$ to end in the digits $00$. In other words, we want $n^2$ to be divisible by $100$ and thus $n$ to be divisible by $10$. The smallest possible solution is $n=20$ and we see that $10\cdot 30+23=323$ does indeed end at $23$. So the smallest possible outcome is $323$.

b. We take again the integers $n-10$ and $n+10$. Now we need to find, for a certain integer $k$, a solution for $n^2-100+23=k^2$, or $n^2=k^2+77$. The difference between two consecutive squares is an odd number that becomes $2$ bigger every time. We have that $1^2-0^2=1$, $2^2-1^2=3$, $3^2-2^2=5$, etcetera. In general: $(m+1{)}^2-m^2=2m+1$. We can get $77$ by taking $m=\frac{76}{2}=38$ and so $k=38$ and $n=39$. We see that indeed it holds that $29\cdot 49+23=1444=38^2$. So it turns out to be possible that the result is a square. In fact, it turns out that this solution is unique, but the problem did not ask us to prove that.