
An interesting problem was posed on algebraonline.com in the beginning of September, 2000:
My initial efforts were along these lines:
Let a, b, and c be the lengths of the sides of a right triangle with integer sides. These numbers, a, b, and c, are called "Pythagorean Triples". All Pythagorean Triples can be found using the following formula:
a = d*(m²  n²),
b = 2dmn,
c = d*(m² + n²),
where d is any positive integer, m>n>0, and m and n are relatively prime and of opposite parity (i.e. m+n is odd).
Then the area is ab/2, which is d²(m²n²)(mn), which is d²(m+n)(mn)(m)(n)
d²(m+n)(mn)(m)(n) is a perfect square if and only if (m+n)(mn)(m)(n) is a perfect square. (because only perfect squares have rational square roots) In other words, dividing by d² doesn't change whether the number is a perfect square. So to show the area, ab/2 is not a perfect square, I need only show that (m+n)(mn)(m)(n) is not a perfect square.
GCD(m,n)=GCD(m+kn,n), where k is an integer. (Linear Combination)
Since GCD(m,n)=1, then m is relatively prime to m+n and mn. Similarly, n is relatively prime to m+n and mn.
GCD(m+n,mn) = GCD(2m,mn) is at most 2, but since mn is an odd number, GCD(m+n,mn)=1.
Now I have shown that every factor is relatively prime to every other factor.
So either each of the four factors is a perfect square, or (m+n)(mn)(m)(n) is not a perfect square.
Now I'm stumped again. I would like to proceed this way: If mn, m, and m+n are all perfect squares, then n is a multiple of 24 (Here's the proof of this). Since n is a multiple of 24 and a perfect square, it must be a multiple of 144. But I don't see how this helps me...
After asking for help on nrich.maths.org, Michael Dor� came to my rescue with this message:
I think you're going about it the right way, but are going to need some sort of descent argument to finish it off. It's probably easier not to apply the Pythagorean formula straight away, but first deduce facts about the sides.
OK, so we have a,b,c sides of a triangle and ab/2 is a square. Let us take c to be minimal  so in other words, we're looking for the counterexample with the smallest value of c possible. (This is possible because c is a natural number.)
Now a,b,c are pairwise coprime (if two of them share a prime factor p then it is easy to check that the third also shares this factor, so (a/p),(b/p),(c/p) form a smaller counterexample; this contradicts the minimality of c).
Also exactly one of a,b is even (otherwise c² = a� + b� = 2 (mod 4)) and c is odd. So without loss of generality we can set a = 2z.
The primitive Pythagorean triple (a,b,c) becomes (2z,b,c), and zb is a square.
Therefore 4z² + b² = c² where b,c are odd and z,b,c are pairwise coprime. The area is zb which is a square therefore z,b are both squares so z = u², b = v² and we get:
4u^{4} + v^{4} = c²
The primitive Pythagorean triple (2z,b,c) becomes (2u²,v²,c)
Since this is a triple, we can find r,s such that 2u²=2rs, v²=r²s², c=r²+s². (More about finding r and s for Pythagorean triples)
Take a close look at the middle expression, v²=r²s². It reveals the existence of yet another Pythagorean triple.
Another Pythagorean triple is (s,v,r)
(s,v,r) is a primitive Pythagorean triple (if it weren't primitive, u,v wouldn't be coprime). And v is odd so once again, we can find two numbers, x,y such that
s = 2xy,
v = x²  y²,
r = x² + y².
But u² = rs = 2xy(x²+y²) and as x is odd and y is even, 2y,x,x²+y² are pairwise coprime so they are all squares. Since 2y is a square, and y is even then y/2 is also a square, so there exists an integer f such that y/2=f². Since x is a square, there exists an integer g such that x=g², and since x²+y² is also a square, an integer h exists such that x²+y²=h². Summarizing this,
y = 2f²,
x = g²,
x²+y² = h².
Then g^{4} + 4f^{4} = h².
Now consider the triangle with sides g²,2f²,h. This is clearly a rightangled triangle since g^{4} + 4f^{4} = h² and it has area (g²*2f²)/2 which is square.
The last Pythagorean triple is (2f²,g²,h)
But h < h² = x²+y² = r < c. This contradicts the minimality of c, so we're done.
Now, returning to my earlier topic, it is clear that
(m+n)(mn)(m)(n)
can't be a perfect square, because if it were,
a = (m²  n²)
b = 2mn
c = (m² + n²)
would be a Pythagorean Triple with "square area" (i.e. ab/2 is a square) which we now know is false. That means that an arithmetic progression of three squares must have a common difference which is not, itself, a square.
Still to be explored...
Whether this little factoid can be used to prove that an arithmetic progression of four squares cannot exist.
Formulas for Primitive Pythagorean Triples and their deriviation  a way to generate all the triples such that a^2 + b^2 = c^2
Puzzle question  if m is the product of n distinct primes, how many different right triangles with integer sides have a leg of length m?
Arithmetic Sequence of Perfect Squares, page 3  If a^2, b^2, c^2 are in arithmetic sequence, why is their constant difference a multiple of 24? Look at the second answer to this question for the relationship between Pythagorean Triples and this arithmetic sequence of squares.
Theorems Involving Perfect Squares  answers questions such as "why is the square root of x irrational unless x is a perfect square?" and other fundamental questions about perfect squares.
The webmaster and author of this Math Help site is Graeme McRae.