Crux-Lab

Euler 9 — Reducing the triplet search to one parameter

Problem

Find natural numbers a < b < c such that:

• a^2 + b^2 = c^2
• a + b + c = 1000

Compute abc.

Key idea

I started by looking at small Pythagorean triples and noticed I could rewrite the product in the form abc = (a+b+c)cx. That turns the problem into a one-parameter search instead of a 3-variable search.

Reduction

From the setup:

• ab = 1000x
• c = 500 - x
• a + b = 500 + x

So a and b must be roots of:

t^2 - (500 + x)t + 1000x = 0

For integer roots, the discriminant must be a perfect square.

Working value

The valid value is x = 75.

So:

• c = 425
• a + b = 575
• ab = 75000

which gives:

• a = 200
• b = 375

Hence:

(a,b,c) = (200,375,425)

and

abc = 31875000

What I learned

The main missing idea was using Vieta’s formulas plus the discriminant condition:

If two integers have sum M and product N, then they are roots of
t^2 - Mt + N = 0.

So integer-root constraints can finish a problem after algebraic reduction.

Checker

```python for x in range(1, 500): s = 500 + x p = 1000 * x D = ss - 4p r = int(D*0.5) if rr == D: a = (s - r) // 2 b = (s + r) // 2 c = 500 - x if a > 0 and b > 0 and a < b < c: print(x, a, b, c, abc) break

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