Project Euler: Problem 9

This is Problem 9 on Project Euler:

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

a² + b² = c²

For example, 3² + 4² = 9 + 16 = 25 = 5².

There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product abc.

Pythagorean triplets can be generated by using the following formulas, where m and n are integers and m is greater than n:

m² – n²
2mn
m² + n²

So for the purposes of this problem we are looking for values for m and n where these three formulas add up to 1000, that is:

(m² – n²) + (2mn) + (m² + n²) = 1000
This can be simplified to 2m² + 2mn = 1000
Dividing by 2 and factoring by m we have m(m+n) = 500
Therefore m must be a factor of 500. Calculating m, m + n and n we get the following table:

Therefore the only solution is where m = 20 and n = 5. Plugging these values into our original expressions we have:

a = m² – n² = (20 * 20) – (5 * 5) = 375

b = 2mn = 2 * 20 * 5 = 200

c = m² + n² = (20 * 20) + (5 * 5) = 425

a + b + c = 1000 and ² + b² = c², so the answer to the problem is a * b * c.

We have not had to use a computer to solve this problem, but a brute-force approach in Delphi could be this:

procedure TfrmMain.Problem9;
var
	a, b, c : integer;
begin
	for a := 1 to 998 do begin
		for b := a + 1 to 999 do begin
			c := 1000 - (a + b);
			if (a * a) + (b * b) = (c * c) then begin
				ShowMessage(IntToStr(a * b * c));
				exit;
			end;
		end;
	end;
end;

This code just generates sets of numbers where a + b + c = 1000 and then checks the set for being a Pythagorean triple. If it is, the product is displayed. There are no optimizations in this code, which could be improved considerably.