Playing with Primitive Pythagorean Triples!

Number Theory Level 4

Define a Primitive Pythagorean Triple as a set of three positive integers $(a,b,c)$ such that $a<b<c$ and $a^2 + b^2 = c^2$ , and where the integers $a,b,c$ don't all share a common factor greater than 1.

How many Primitive Pythagorean Triples exists that satisfy the equation $a+b-c = 20$ ?

\infty

2 0 6 4 1 5 3

1 solution

Shaun Leong
Dec 29, 2015

We will use the parametrization of Pythagorean Triples, $(a,b,c)=(m^2-n^2,2mn,m^2+n^2)$ . Note that we can swap the values of $a$ and $b$ to satisfy $a <b <c$ since they are symmetric in $a+b-c=20$

Substituting these values into the equation, $m^2-n^2+2mn-(m^2+n^2)=20$ $2mn-2n^2=20$ $n(m-n)=10$

By trying $n=1,2,5,10$ , we obtain $(m,n)=(11,1), (7,2), (7,5), (11,10)$

The corresponding values of $a,b$ and $c$ are $(a,b,c)=(22,120,122), (28,45,53), (24,70,74), (21,220,221)$

The first and third solutions are rejected since their $\gcd > 1$ . This leaves us with $\boxed {2}$ solutions.

Playing with Primitive Pythagorean Triples!

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