Pythagoras' triangles proof

Question

Let $A,B,C$ and $a,b,c$ are the sides of Pythagoras' triangles. Prove that there exists an integer $D$ or $E$ such that $(C+c)^2-(A+a)^2-(B+b)^2=D^2$ or $(C+c)^2-(A+a)^2-(B+b)^2=2E^2$

What is the method to prove this? Is it possible to use primitive Pythagoras' triples?

Answer 1

Assuming that $c$ and $C$ are the hypotenuses, and that both $(a,b,c)$ and $(A,B,C)$ are primitive, then there are integers $p$, $q$, $r$, $s$ such that $$c = p^2+q^2,\ b = p^2-q^2,\ a=2pq, \quad C = r^2+s^2,\ B=r^2-s^2,\ A=2rs$$ or (reversing $a$ and $b$, and $A$ and $B$ if necessary) $$c = p^2+q^2,\ b = p^2-q^2,\ a=2pq, \quad C = r^2+s^2,\ B=2rs,\ A=r^2-s^2.$$ Now, $(C+c)^2-(A+a)^2-(B+b)^2 = 2(Cc-Aa-Bb)$; substituting and simplifying gives for the first case $$2(Cc-Aa-Bb) = 2(qr-ps)^2$$ and for the second $$2(Cc-Aa-Bb) = (p (r-s)-q (r+s))^2.$$

Answer 2

Does the term Pythagoras' triangles imply that the sides are pairwise coprime and the triangle is right? Without pairwise coprime the theorem is certainly not true.

The LHS is equal to $$C^2 - A^2 - B^2 + c^2 - a^2 - b^2 + 2(Cc - Aa - Bb)$$

Take an example where $C, c$ are the hypotenuses. Let $Cc - Aa - Bb = K$. Then the expression is equal to $2K$, suppose that this was either a perfect square or twice a perfect square. But there must be some prime $p$ such that $p \nmid 2K$, and if we take $A' = pA, B' = pB, C' = pC$ then the theorem doesn't hold for $A', B', C', a, b, c$.

Answer 3

Hint:

By ''Pythagoras triangle'' I suppose that you means that $A,B,C$ and $a,b,c$ are Pythagorean triples, such that:

$$ C=M^2+N^2 \qquad B=2MN \qquad A=M^2-N^2 $$ and

$$ c=m^2+n^2 \qquad b=2mn \qquad a=m^2-n^2 $$

for $M,N , m, n$ integers.

Substitute these in your equation and you can find that $$ D^2=4(Mn+Nm)^2 $$