Suppose that a is even and b c are odd Prove that a b c is a

Suppose that a is even, and b, c are odd. Prove that (a, b, c) is a primitive Pythagorean triple i there exist even s and odd t
with (s, t) = 1 such that a = 2st, b = s^2 t^2, and c = s^2 + t^2.

Solution

if a = 2st

b = s^2 - t^2

c = s^2 +t^2

with (s,t) = 1

and s even , t odd

then

a^2 + b^2 = c^2 , a is even , b is odd , c is odd

also (a,b,c) is primitive

they don\'t have common divisor.

suppose they have a common prime divisior , d (d is not 2) then

d divides b , d divides c

d divides b + c = 2s^2 => d divides s^2 => d divides s

d divides c - b = 2t^2 => d divides t^2 => d divides t

gcd(s,t) not equal to 1

this is the contradiction

So (a,b,c) cannot have common divisor

Now suppose (a,b,c) are primitive triple

a even , b, c are odd

then

a^2 = c^2 - b^2

a^2 = (c-b) (c+b)

both c-b and c+b are even

so

a^2/4 = (c-b)/2 (c+b)/2

claim

(c-b)/2 and (c+b)/2 are coprime

if there is some common prime divisor p then

p divides c( (c-b)/2 +(c+b)/2 ) and p divides b ( (c+b)/2 -(c-b)/2 ) and so p divides (c^2 - b^2 = a^2)

which will make (a,b,c) non primitive

so (c+b)/2 and (c-b)/2 are coprime.

also (c+b)/2 (c-b)/2 = (a/2)^2

this means (c+b)/2 is complete sqaure and (c-b)/2 is complete sqaure

(c+b)/2 = s^2

(c-b)/2 = t^2

a/2 = st => a = 2st

since s^2 + t^2 = c odd

so one of s or t is odd and other even

there is an error in the question s need not be even, t can be even and s odd for example

pair (12,5,13)