A square has side length $s$ and a diagonal of length $s+1$. Write the area of the square in the form $a+b \sqrt{c}$ where $a, b$ and $c$ are positive integers.
If a square has side length $s$, then the length of its diagonal is $\sqrt{2} s$.

Since we are told that the length of the diagonal is $s+1$, then $\sqrt{2} s=s+1$ or $\sqrt{2} s-s=1$ or $(\sqrt{2}-1) s=1$.

Thus, $s=\frac{1}{\sqrt{2}-1}=\frac{\sqrt{2}+1}{(\sqrt{2}-1)(\sqrt{2}+1)}=\frac{\sqrt{2}+1}{2-1}=\sqrt{2}+1$.

The area of the square is $s^{2}=(\sqrt{2}+1)^{2}=2+2 \sqrt{2}+1=3+2 \sqrt{2}$.

ANSwER: $3+2 \sqrt{2}$