Finding a Pythagorean triple $a^2 + b^2 = c^2$ with $a+b+c=40$
Assuming you do have a pen and paper, you could substitute $c = 40 - a - b$ into the first equation to get
$$a^2 + b^2 = (40 - a - b)^2 = a^2 + b^2 + 1600 - 80(a + b) + 2ab.$$
Rewriting this equation, you get
$$a + b - 20 = \frac{ab}{40}.$$
From this it follows that $ab$ has to be a multiple of $40$, i.e., one of them is a multiple of $5$. That narrows it down to only a few options...
If that's still too much brute-force, you could also note that $a + b > 20$ from the above equation, and $a + b < 27$, since $c$ has to be the largest of the three. This leaves only the three pairs
$$\{(5,16),(10,16),(15,8)\}.$$
Looking at the earlier equation, you see the third pair is the right one.
The general pythagorean triple can be written (up to swapping $a$ and $b$) as $$a=2kuv$$ $$b=k(u^2-v^2)$$ $$ c=k(u^2+v^2)$$ where $k,u,v$ are positive integers, $u,v$ are relatively prime with different parity and $u\geq v$. For $a+b+c=40$, then, you get the condition $2k(u^2+uv)=40$, so you need $u^2+uv=u(u+v)$ to be a factor of $20$. Since $u,v$ have different parity, and they are positive, you know $u+v>1$ is odd, so $u+v=5$.
Given that $u\geq v$, that yields $u=4,v=1$ and $u=3,v=2$. But $u=3$ isn't possible, since $3$ is not a factor of $20$. So The only solution is $(u,v)=(4,1)$ and therefore the only solution is $k(15,8,17)$ which you can see must have $k=1$, and you are done - the only solution is $(15,8,17)$. And $(8,15,17)$, if you count that as different.
For the more general problem, $a+b+c=2n$ (the sum of a Pythagorean triple is always even) this amounts to factoring $n=kuw$ with the following conditions:
- $k,u,w$ are positive in
- $w$ is odd
- $u<w<2u$
Given such a solution, you get a triple (by setting $v=w-u$:) $$a=2ku(w-u)$$ $$b=kw(2u-w)$$ $$c=k(2u^2+w^2-2uw)$$
And this gives all such triples (modulo swapping $a$ and $b$.)
Listing these amounts to first listing the set of possible values of $w$, which can be any odd factors of $n$ such that $1<w<\sqrt{2n}$. Then find values of $u$ with $w/2<u<w$ and $uw|n$. Then set $k=n/uw$ and you have your triple $(k,u,w)$ from which you can compute an $(a,b,c)$.
(If you want to allow $ab=0$, the change the above to $u\leq w < 2u$. Then $u=v=1$ and $k=n$ is always a solution.)
Neat question. If one were to allow for answers in $\mathbb{Z}$, you can have as many as 18 solutions. $$\begin{align} &(65,72,-97) &&(90,56,-106) &&&(140,48, -148) \\ &(240,44,-244) &&(440, 42, -442) &&&(840, 41, -841) \\ &(45,200,-205) &&(50,120,-130) &&&(60,80,-100) \end{align}$$
$$\begin{align} &***(8,15,17)*** &&(24,-10, 26) &&&(32,-60,68) \\ &(36,-160,164) &&(38,-360,362) &&&(39,-760,761) \\ &(-120,35,125) &&(-40,30,50) &&&(0,20,20) \end{align}$$
Solution: $$\begin{cases} a^2+b^2=c^2 &&&&&&(1)\\a+b+c=40 &&&&&&(2)\end{cases}$$
$$(2) \ \text{for c into} \ (1)\to $$ $$\begin{align} a^2+b^2&=[40-(a+b)]^2 &&\implies \\ a^2+b^2&=1600-80(a+b)+(a+b)^2 &&\implies \\ 0&=1600-80(a+b)+2ab &&\implies \\ 0&=ab-40(a+b)+800 &&\implies \\ 800 &= ab - 40(a+b) + 1600 &&\implies \\ 800 = d_1 \cdot d_2&= (a-40)(b-40)&&\implies \\ &\Bigg{\Downarrow} \\ \end{align}$$
$$\begin{cases} a=40+d_1 \\ b=40+d_2 \\ c =-40-(d_1+d_2)\end{cases} $$
The sides of the triangle are now defined parametrically in terms of an arbitrary choice of divisors of $800$. Here, for any choice of a divisor-pair $(d_1,d_2)$, switching the order doesn't produce a new solution, (other times it can), but there is an additional $(-d_1,-d_2)$ solution.
Since, $$\begin{align} 800=2^{\color{red}{5}} \cdot 5^{\color{red}{2}} &\implies (\color{red}{5}+1)(\color{red}{2}+1)=18 \ \text{divisors} \\ &\implies 9 \ \text{divisor-pairs} \\ &\implies 18 \ \text{divisor-pairs including negatives} \end{align}$$
we expect $18$ solutions overall, evaluated above.