Evaluating $\sqrt{119^2+120^2}$ with clever algebra Numbers $(119,120,169)$ are Pythagorean triples, i.e $119^2+120^2=169^2$. I'm wondering is it possible to start from $119^2+120^2$ and get $169^2$ algebraically without evaluating $119^2$  and $120^2$ directly? I could guess $169$ with approximation,
$$\sqrt{119^2+120^2}\approx\sqrt{2\times 120^2}=120\sqrt2\approx120\times1.4$$Which gives $168$ and checking the unit digits $169$ is a reasonable guess and it works!
But I don't know if it is possible to start with $119^2+120^2$ and get $169^2$.

Whatever $119^2 + 120^2$ equals, it does end in a $1$. This means that if $c = \sqrt{119^2+120^2}$ is an integer, that integer ends in $9$ or $1$. Now let's try to put bounds on what this integer could be. In this case, because the numbers are so close to each other, we can immediately put a fairly tight bound on $c$: $119\sqrt{2} < c < 120\sqrt{2}$. This means we can use a rather loose rational approximation of $\sqrt{2}$, such as $\sqrt{2} > 7/5$, and still get decent bounds. A quick calculation then shows $166 < c < 172$. Since $171$ is divisible by $3$ and $c$ clearly isn't, that means if $c$ is an integer, it must be $169$.
For general $c = \sqrt{a^2+b^2}$, looking for a way to make approximate bounds on $\sqrt{a^2+b^2}$, then eliminating numbers in that interval using divisibility and modular arithmetic will often suffice to reach one or zero candidates for integer $c$. At that point you can check your single candidate using pythagorean triple formulae, or stop if the set of candidates is empty.
