[Number Theory]
Problem: Compute the number of positive integers less than 25 that cannot be written as the difference of two squares of integers.

Solution: Suppose $n=a^{2}-b^{2}=(a+b)(a-b)$, where $a$ and $b$ are integers. Because $a+b$ and $a-b$ differ by an even number, they have the same parity. Thus $n$ must be expressible as the product of two even integers or two odd integers. This condition is sufficient for $n$ to be a difference of squares, because if $n$ is odd, then $n=(k+1)^{2}-k^{2}=(2 k+1) \cdot 1$ for some integer $k$, and if $n$ is a multiple of 4 , then $n=(k+1)^{2}-(k-1)^{2}=2 k \cdot 2$ for some integer $k$. Therefore any integer of the form $4 k+2$ for integral $k$ cannot be expressed as the difference of two squares of integers, hence the desired integers in the given range are $2,6,10,14,18$, and 22 , for a total of 6 values.
Suppose that an integer $n$ can be expressed as the difference of squares of two integers, and let the squares be $a^{2}$ and $(a+b)^{2}$, with $a, b \geq 0$. Then

$$
\begin{aligned}
& n=(a+b)^{2}-a^{2}=2 a b+b^{2} \\
& =2 a+1 \quad(b=1) \\
& =4 a+4 \quad(b=2) \\
& =6 a+9 \quad(b=3) \\
& =8 a+16 \quad(b=4) \\
& =10 a+25 \quad(b=5) .
\end{aligned}
$$

Setting $b=1$ generates all odd integers. If $b=3$ or $b=5$, then the values of $n$ are still odd, hence are already accounted for. If $b=2$, then the values of $4 a+4=4(a+1)$ yield all multiples of $4 ; b=8$ yields multiples of 8 (hence are already accounted for). The remaining integers are even numbers that are not multiples of $4: 2,6,10,14,18,22$, for a total of 6 such numbers.

Final answer: 6