Find the number of positive integers less than 101 that can not be...

Let, $\mathrm{c} \rightarrow$ be positive integer $\leq 100$ such that $\mathrm{c}^2=\mathrm{a}^2-\mathrm{b}^2 ; \mathrm{a}, \mathrm{b} \in \mathrm{Z}$. $\mathbf{C}-\mathbf{I}$ : Difference of a and b is 1 . $$\begin{array}{ll}\therefore & 1^2-0^2=1 \\ & 2^2-1^2=3 \\ & 3^2-2^2=5 \\ & 4^2-3^2=7 \\ & 5^2-4^2=9 \\ & \vdots \\ & 50^2-49^2=99\end{array}$$ All odd number upto 100 can be expressed as difference of two squares. $\mathbf{C}-\boldsymbol{I I}$ : Difference of $\mathrm{a}, \mathrm{b}$ is 2 $$\begin{aligned} & 2^2-0^2=4 \\ & 3^2-1^2=8 \\ & 4^2-2^2=12 \\ & \vdots \\ & 26^2-24^2=100\end{aligned}$$ All multiples of ' 4 ' upto 100 can be expressed as difference of two squares. $\mathbf{C}$ - III : Difference of $\mathrm{a}, \mathrm{b}$ is 3 This case will give odd numbers which are already counted $\mathbf{C}$ - IV : Difference of $\mathrm{a}, \mathrm{b}$ is 4 This case will give multiples of 4 which are already counted Similarly, the remaining cases will all give number already counted in $\mathrm{C}-\mathrm{I}$ and $\mathrm{C}-\mathrm{II}$ $\therefore \quad$ No. of numbers which can be expressed as difference of squares $=($ odd numbers $)+($ multiples of 4$)=50+25=75$ $\therefore \quad$ Required numbers which cannot be expressed as a difference of squares of two integers are $=100-75=25$