C2. Two positive integers with a difference of 20 are multiplied together; then 23 is added to the result.

(a) What is the smallest possible outcome that ends in 23? Provide this outcome (and the two corresponding numbers with a difference of 20) and further prove that there is no smaller outcome that ends in 23.

(b) Is it possible for the outcome to be the square of an integer? Provide an example (and show that it satisfies the condition) or prove that this is not possible.
C2. (a) Suppose the two positive integers are $n-10$ and $n+10$, and thus $n>10$. Then the product is $(n-10)(n+10)=n^{2}-100$, and we are looking for an $n>10$ such that $n^{2}-100+23$ ends in the digits 23. This 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 $n$ must be divisible by 10. The smallest possible solution is $n=20$, and we see that $10 \cdot 30+23=323$ indeed ends in 23. The smallest possible result is therefore 323.

(b) We take the numbers $n-10$ and $n+10$ again. Now we are looking for a solution to $n^{2}-100+23=k^{2}$ for some integer $k$, or in other words, $n^{2}=k^{2}+77$. The difference between two consecutive squares is an odd number, and this difference increases by 2 each time. It holds that $1^{2}-0^{2}=1, 2^{2}-1^{2}=3, 3^{2}-2^{2}=5$, and so on. In general: $(m+1)^{2}-m^{2}=2m+1$. We can get 77 by taking $m=\frac{76}{2}=38$, and thus $k=38$ and $n=39$. We see that indeed $29 \cdot 49+23=1444=38^{2}$. It turns out that it is possible for the result to be a square. It turns out that this is the only solution, but that is not asked for in the problem.