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The "Euler Project" problem 9 (https://projecteuler.net/problem=9) asks to solve:

$a^2$ + $b^2$ = $c^2$

a + b + c = 1000

I find answers solving it with brute-force and programmatically, but is there a way to solve the problem ONLY mathematically? Can someone help, please?

Problem as explained in Project Euler website:

A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,

a^2 + b^2 = c^2
For example, 3^2 + 4^2 = 9 + 16 = 25 = 52.

There exists exactly one Pythagorean triplet for which a + b + c = 1000.

Find the product abc.

Hint The classical parameterization of the Pythagorean triples is
$$a = k (m^2 - n^2), qquad b = 2 k m n, qquad c = k (m^2 + n^2),$$
where $m > n > 0$ and $m, n$ coprime and not both odd.

Substituting in our condition gives
$$1000 = a + b + c = 2 k m (m + n),$$ and clearing the constant leaves $$phantom(ast) qquad 500 = k m (m + n) . qquad (ast)$$
Now, notice that (1) $500 = 2^2 5^3$ has only two distinct prime factors, and (2) since $m$ and $n$ are coprime, so are $m$ and $m + n$.

So, one of $m, m + n$ must be one of $1, 2, 4$ (in fact one of $2, 4$, since $m > n > 0$ implies $m + n > m > 1$) and the other must be one of $1, 5, 25, 125$. Since $m + n > m$, we must have $m in 2, 4$, and so $m + n < 2 m leq 8$. Thus, $m + n = 5$, and $2 m > m + n = 5$ implies $m geq 3$, leaving $m = 4$ as the only possibility. Substituting gives $n = 1, k = 25$ and so substituting in our parameterization we find $$color#df0000boxed(a, b, c) = (275, 200, 425) .$$

It is known that all primitive Pythagorean triples (i.e with no common factors) is of the form $(m^2-n^2,2mn,m^2+n^2)$ for relatively prime $m$ and $n$ where one is even and the other is odd.

Based on that, you are looking for $m$ and $n$ such that $$(m^2-n^2)+2mn+(m^2+n^2)=2m^2+2mn=2m(m+n)$$ is a factor of $1000$.

The triplets are all of the form
$a=u(n^2-m^2),
b=2umn,
c=u(n^2+m^2)
$
with $n > m$
so
$a+b+c
=u(2n^2+2mn)
=2un(n+m)
$.

We must have
$n > m$.

Therefore
$500
=un(n+m)
$.

If
$500 = rst
$
with
$s < t$
then
$u = r,
n = s,
n+m = t
$
so
$m = t-n
=t-s
$.

We must have
$n > m$
so
$s > t-s$
or
$s < t < 2s$.

Playing around a bit,

$500 = 1*20*25$,
so,
swapping $m$ and $n$,
$u = 1, m = 5,
n=20
$
and the sides are
$20^2-5^2 = 375 = 25 15,
2 20 5 = 200 = 25 8,
20^2+5^2 = 425 = 25 17
$.

Given perimeter: $qquad P=(m^2-n^2 )+2mn+(m^2+n^2 )=2m^2+2mnqquad $ If we solve for $n$, we can find if there exists one or more $m,n$ combinations for a Pythagorean triple with that perimeter. Any value of $m$ that yields an integer $n$ gives us such an $m,n$ combination. We let:

$$n=fracP-2m^22mquad where quad biggllceilfracsqrtP2biggrrceille m le biggllfloorsqrtfracP2biggrrfloor$$
Here, the lower limit ensures that $m>n$ and the upper limit insures that $n>0$. For example:

$$P=1000implies biggllceilfracsqrt10002biggrrceil =16le m le biggllfloorsqrtfrac10002biggrrfloor=22$$

In this range, we find only one value of $m$ that yields and integer $n$: $m=20implies n=5$, and, using Euclid's formula $F(m,n)$, we have $F(20,5)=(375,200,425)$. Then, the product, as I understand it, is $$Atimes Btimes C=375times200times425=31875000.$$