Question about Pythagorean triples Given $a$, $b$, $d$ natural numbers.
Suppose $(a, b)$ are two legs of a Pythagorean triple.
Also suppose $(a, b+d)$ are two legs of another Pythagorean triple.
I'm looking for a way to show that given the above:
$(a+d,b)$ cannot be the legs of a Pythagorean triple. (I don't know if this is true... But I'd like to get some ideas)
Would appreciate any tips on how to attack such a problem (not a homework problem, just my own personal interest). 
 A: Primitive pythagorean triples are generated by positive integers $p,q$, where $\gcd{(p,q)}=1$ and $p,q$ are of opposite parity. In particular, if $(a,b,c)$ form a primitive pythagorean triple, then we must have $a=2pq, b=p^2-q^2$ and $c=p^2+q^2$. 
For example, if $p=1, q=2$, then $a=2\cdot 2\cdot 1=4, b=2^2-1^2=3$ and $c^2=2^2+1^2=5$.  Thus, you have your $(3,4,5)$ we all know and love.
Non primitive pythagorean triples are just scalar multiples of pythagorean triples. Perhaps you can use these facts.  You can say, suppose $a,b$ are two legs of a pythagorean triple.  Then $a=2p_1q_1, b=p_1^2-q_1^2$ is it is a primitive pythagorean triple or $a=k(2p_2q_2), b=k(p_2^2-q_2^2)$ is any pythagorean triple.   Now how do you consider $b+d$? 
A: Problems like this can often be solve by scaling.  Your two larger triangles have legs that sum to the same total.  If we start with $3,4,5$ and $5,12,13$ we can multiply the first by $17$ and the second by $7$ giving $51,68,85$ and $35,84,91$ as two triples with common leg sum.  Unfortunately, $35,68$ don't make two legs of a triangle because $35^2+68^2=5849=11\cdot 499$  You can search over pairs of primitive triangles this way looking for a solution.
