Question

explain Pythagorean triple (m²-n²),2mn,(m²+n²)​
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Answer 1

Proof.

If (a, b, c) is a primitive Pythagorean triple with b even, then a is odd, and since c ^2 = a ^2 + b ^2 ,

then c is also odd. Therefore, c − a and c + a are both even,

so that

(c + a )/2 · (c − a) /2 . = (b/2)^2 .

Any common divisor of the two factors on the left divides both their sum c and their difference a, and

since gcd(a, c) = 1, then the two factors on the left have no common divisors except 1, that is, they

are relatively prime. From the lemma, they must both be squares, so there exist positive integers m

and n such that

(c + a )/2 = m^2

(c + a )/2 = n ^2

b/ 2 = mn.

We note that m and n are relatively prime since c + a and c − a are relatively prime, and that m > n.

Also,

since c ^2 = m^2 + n^ 2 and c is odd, then m and n have different parity.