For example:
- two consecutive integers $3$ and $4$, with $\sqrt{3^2+4^2}=5$
- and for $20$ and $21$ answer the square root of $20^2+21^2$ is $29$
- for $0$ and $1$ the square root is $1$.
Any other examples?
For example:
Any other examples?
Consider the sequence $(a_n)$ given by $$ a_0=0, \qquad a_1=3, \qquad a_{n}=6a_{n-1}-a_{n-2}+2\text{ for }n\ge 2$$ and the sequence $(c_n)$ given by $$ c_0=1, \qquad c_1=5, \qquad c_{n}=6c_{n-1}-c_{n-2}\text{ for }n\ge 2.$$ Then $a_n^2+(a_n+1)^2=c_n^2$ is immediately verified for $n=0$ and $n=1$. Then by induction these can be expressed in terms of $\alpha=3+2\sqrt 2$ and $\beta=3-2\sqrt 2$. More precisely, we verify that $$ a_n = \frac{1+\sqrt 2}{4}\alpha^n+\frac{1-\sqrt 2}4\beta^n-\frac12$$ $$ c_n = \frac{2+\sqrt 2}{4}\alpha^n+\frac{2-\sqrt 2}{4}\beta^n$$ fulfill both the initial conditions and the recursion hence describe our sequencs explcitly. Then (using $\alpha\beta=1$) $$ a_n^2= \frac{(1+\sqrt 2)^2}{16}\alpha^{2n}+\frac{(1-\sqrt 2)^2}{16}\beta^{2n}+\frac18-\frac{1+\sqrt 2}{4}\alpha^n-\frac{1-\sqrt 2}4\beta^n$$ $$ (a_n+1)^2= \frac{(1+\sqrt 2)^2}{16}\alpha^{2n}+\frac{(1-\sqrt 2)^2}{16}\beta^{2n}+\frac18+\frac{1+\sqrt 2}{4}\alpha^n+\frac{1-\sqrt 2}4\beta^n$$ $$ c_n^2=\frac{(2+\sqrt 2)^2}{16}\alpha^{2n}+\frac{(2-\sqrt 2)^2}{16}\beta^{2n}+\frac14$$ so that by a lot of cancellation and surd simplification $c_n^2=a_n^2+(a_n+1)^2$ for all $n$.
The first few solutins thus found are $0^1+1^2=1^2$, $3^2+4^2=5^2$, $20^2+2^ 2=29^2$, $119^2+120^2=169^2$, $696^2+697^2=985^2$, $4059^2+4060^2=5741^2$, and can be continued indefinitely. As a matter of fact, the list thus obtained is complete, i.e. there are no other solutions. But to show that would involve more than the above straightforward veififcations.
You are asking for the equation for generating Pythagorean triples: $$a = x^2 - y^2 ,\ \, b = 2xy ,\ \, c = x^2 + y^2$$ for any two positive integers $x$ and $y$ with $x > y$. For consecutivity, $ x^2-y^2=2xy\pm1 $ is a requirement for the legs to have consecutive integer side lengths.