Problem 2003

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.