Well well, this was so far the easiest problem of the lot. The requirement is to find a Pythagorean triplet a, b, c such that
a + b + c = 1000
I was able to solve this question using pencil and paper.
A primitive Pythagorean triplet is one for which:
a^2 + b^2 = c^2;
gcd(a,b) = gcd(b,c) = gcd(a,c) = 1
Such a triplet can be written in the following parametrized form:
a = m^2 – n^2
b = 2mn
c = m^2 + n^2
for integers m, n. Using the parametrized form for a, b, c and the condition of their sum being 1000, we get the following:
a + b + c = 1000
=> 2m^2 + 2mn = 1000
=> m^2 + mn – 500 = 0
Roots of this equation can be easily found as m = 20 and n = 5. Thus we have that
a = 375, b = 200 and c = 425
And our problem is solved.
Now what if we replace 1000 by s. In that case the quadratic in m changes and we will have to code a small loop to determine an integral value for m, n. But even that is straight forward and I see not difficulty in it.
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