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 $169^2$.
 A: 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.
A:  Minor note: This answer doesn't care where the numbers come from. It simply obtains whether this expression is a perfect square without using a numerical approximation.

Suppose that $119^2+120^2=a^2$, where $a\in \mathbb N.$
Let $m\in\mathbb N$, then we have:
$$
\begin{aligned}&(a-120)(a+120)=119^2\\
\implies &\begin{cases} a-120=m
\\a+120=\frac {119^2}{m}\end{cases}\\
\implies & m<119\wedge m=7^x\cdot 17^y\\
\implies &\frac {119^2}{m}-m=240,\, m>1\\
\implies &7^{2-x}\cdot 17^{2-y}-7^x\cdot 17^y=2^4\cdot 15\\
\implies &x=2\wedge y=0\\
\implies &m=7^2\\
\implies &a=7^2+120\\
&\,\,\, =17^2-120\\
&\,\,\, =169.\end{aligned}
$$

*

*We used the fact:

$$240\not\equiv 0\pmod {7,17}.$$
A: Once  you know it's a valid Pythagorean triple, you can do things like:
$$119^2 + 120^2 = (12^2-5^2)^2 + (2\cdot 12\cdot 5)^2 = 12^4+2\cdot 12^2\cdot 5^2 + 5^4 = (12^2+5^2)^2 = 169^2.$$
But this is basically cheating and I can't think of a canonical way to do this if you didn't know it was a whole square to begin with. For determining $12$ and $5$, you just remember the formula $(m^2-n^2, 2mn, m^2+n^2)$ for all primitive Pythagorean triple.
A: Here is a more indirect approach than what's been suggested so far, which has the benefit of generalizing substantially. Let's try to search for Pythagorean triples of the form $(a-1, a, b)$, that is, solutions to
$$a^2 + (a - 1)^2 = 2a^2 - 2a + 1 = b^2.$$
Multiplying by $2$ and rearranging a bit gives
$$(2a - 1)^2 + 1 = 2b^2$$
which is a Pell equation. This one is particularly classical and has known solutions in terms of the Pell numbers $P_n$, which are the sequence of denominators in the convergents of the continued fraction of $\sqrt{2}$; specifically, as it turns out, $b$ must be given by a Pell number $P_{2n+1}$ with odd index. The Pell numbers satisfy $P_0 = 0, P_1 = 1, P_{n+2} = 2P_{n+1} + P_n$ so it's not hard to calculate the first few by hand, and we get the sequence (A000129 on the OEIS)
$$0, 1, 2, 5, 12, 29, 70, 169, \dots $$
with odd terms
$$1, 5, 29, 169, \dots $$
and there's our $169$ in particular, corresponding to $\sqrt{2} \approx \frac{239}{169} = \frac{119 + 120}{169}$. That entry of $29$ corresponds to the triple $(20, 21, 29)$ and the next one turns out to be $(696, 697, 985)$.
I don't think it's possible to do much better than this: since $119$ and $120$ are so close, the LHS is close to $2 \cdot 120^2$ which we want to also be a square, so this question is inevitably closely tied up with the question of rational approximations to $\sqrt{2}$. There is not going to be a neat polynomial identity involved here (beyond the usual $(2mn, m^2 - n^2, m^2 + n^2)$ stuff as has already been noted).
A: Just for the fun
$$A=\sqrt{119^2+120^2}=\sqrt{(121-2\epsilon)^2+(121-\epsilon)^2}$$Expand as series around $\epsilon=0$
$$A=121 \sqrt{2}-\frac{3 \epsilon}{\sqrt{2}}+\frac{\epsilon^2}{484 \sqrt{2}}+\frac{3 \epsilon^3}{117128 \sqrt{2}}+\frac{35 \epsilon^4}{113379904 \sqrt{2}}+\frac{9 \epsilon^5}{2494357888 \sqrt{2}}+O\left(e^6\right)$$ Make $\epsilon=1$
$$A=121 \sqrt{2}-\frac{7477855365}{2494357888 \sqrt{2}}=168.999999999970\cdots$$
A: Show that $119^2 + 120^2 = 169^2$ using algebra.
By observing that
$119^2 = (169 - 50)^2$
$120^2 = (169 - 49)^2$
it becomes imperative to consider the polynomial equation
$\;  (x - 50)^2 +   (x - 49)^2 = x^2$
Putting into standard form,
$\;p(x) = x^2 -198 x + 4901$
we can check that, indeed, $p(169) = 0$.
