<s>[INST] Positive integers $a,b,$ and $c$ satisfy the equations. Find the value of $a+b+c$ 
Question: Positive integers $a,b,$ and $c$ satisfy the equations$$29^2-a^2=28^2-b^2=27^2-c^2$$If $a<20$, what is the value of $a+b+c$?

I first tried factoring the three equations so that$$(29-a)(29+a)=(28-b)(28+b)=(27-c)(27+c)$$But I don't really see a way to continue from there. I then looked at the difference of two squares and drew three right triangles of hypotenuse $29,28,27$ and a leg of length $a,b,c$ respectively.
The third side is then the square root of what we want, but I'm not sure what value still goes into $a$ and $b$ and $c$.
Any ideas on where I can begin?
 [/INST] Take $a$ and $b$ first, we will have:
$a^2-b^2=29^2-28^2=57 \Rightarrow (a-b)(a+b)=57$
Because $57>0, a>0, b>0$, we will have $0<a-b<a+b<57$.
There are only two positive cases: 
$57=1 \times 57= 3 \times 19$.
Because $a-b<a+b$, we will have two set of equations:


*

*Case one: ${\begin{cases}a-b=1\\a+b=57\end{cases}}\Leftrightarrow {\begin{cases}a=29\\b=28\end{cases}}$, this result is eliminated because $a<20$.

*Case two: ${\begin{cases}a-b=...\\a+b=...\end{cases}}\Leftrightarrow {\begin{cases}a=...\\b=...\end{cases}}$
With $a=...;b=...$, you can solve for $c$, but I won't give it away (or you can see the spoiler below).

 $a=11;b=8;c=3\Rightarrow a+b+c=11+8+3=22$

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