Project Euler is a series of challenging mathematical/computer programming problems that will require more than just mathematical insights to solve. Although mathematics will help you arrive at elegant and efficient methods, the use of a computer and programming skills will be required to solve most problems.
The motivation for starting Project Euler, and its continuation, is to provide a platform for the inquiring mind to delve into unfamiliar areas and learn new concepts in a fun and recreational context.
Today, I would like to discuss problem 142. I've seen a post from Santiago Alessandri, so I liked to do the task by myself.
The task is:
Find the smallest x + y + z with integers \(x > y > z > 0\) such that x + y, x  y, x + z, x  z, y + z, y  z are all perfect squares.
I don't want to post the solution (if you want to cheat, I guess you could easily Google it), but some thoughts that might help you to get in the right direction.
First thought: Bruteforce
Bruteforce is the easiest way that could give you the solution. So I wrote this piece of code:
#!/usr/bin/python
# * coding: utf8 *
import sys
from math import sqrt
def is_square(integer):
root = sqrt(integer)
if int(root + 0.5) ** 2 == integer:
return True
else:
return False
for x in xrange(3,1000):
print x
for y in xrange(2, x):
for z in xrange(1, y):
if (x > y and y > z):
if (is_square(x + y)
and is_square(x  y)
and is_square(x + z)
and is_square(x  z)
and is_square(y + z)
and is_square(y  z)):
print ("%i  %i  %i" % (x, y, z))
sys.exit()
This is quite fast until you reach about 500. So this is not a good way to solve it.
Apply some math
You can formalize the task like this: Find the smallest \(x, y, z \in \mathbb{N}\), so that:
 \(x > y > z > 0\)
 \(a = x + y\)
 \(b = x  y\)
 \(c = x + z\)
 \(d = x  z\)
 \(e = y + z\)
 \(f = y  z\)
With \(a, b, c, d, e, f \in Squares\).
Now you can make the following conclusions:
 \(\overset{A.1, A.2, A.3}{\implies} a > b\)
 \(\overset{A.1, A.4, A.5}{\implies} c > d\)
 \(\overset{A.1, A.6, A.7}{\implies} e > f\)

\(a > c\): \(y > z\) \(\Leftrightarrow x + y > x + z\) \(\Leftrightarrow a > c\)

\(c > e\): \(x > y\) \(\Leftrightarrow x + z > y + z\) \(c > e\)

a is the biggest element (see B.1, B.4, B.6)

\(b < c\): \(y < z\) \(\Leftrightarrow x  y < x + z\) \(b < c\)

c is the second biggest element (see B.7, B.2, B.5, B.8)

\(b < d\): \( y > z\) \(\Leftrightarrow y < z\) \(\Leftrightarrow x  y < x  z\) \(b < d\)

\(d > f\): \( x > y\) \(\Leftrightarrow x  z > y  z\) \(d > f\)

I can't tell anything about the relationship between:
 d and e
 b and f
 b and e
Lets conclude:
You also know:
\(x = \frac{a  b}{2} \implies \text{ (a  b) has to be even} \implies \text{a and b have the same parity.}\) The same argumentation can be used for (c, d) and (e, f).
\(x > y > z > 0 \land a = x + y \implies a \geq 5\).
With this in mind you don't have to loop over three variables but only over two. This is much faster. As z is over 1000 you need it. My new script took about 1.5 minutes.
Material
Some material like the LaTeXfile can be found in the Project Euler 142 Archive.