Pythagorean triple. 1. x = p^2-q^2 2. y = 2pq 3. z = p^2 + q^2 k = p^2 and l = q
ID: 3146766 • Letter: P
Question
Pythagorean triple. 1. x = p^2-q^2 2. y = 2pq 3. z = p^2 + q^2k = p^2 and l = q^2. Check that the Pythagorean triple (x, y, z) is given by equation (1, 2, 3). Also show that 1 is the largest common factor for p and q and that they are not both odd numbers.
Assume (x,y,z) is a primitive Pythagorean triple, where x is odd, y is even and z is odd. Pythagorean triple. 1. x = p^2-q^2 2. y = 2pq 3. z = p^2 + q^2
k = p^2 and l = q^2. Check that the Pythagorean triple (x, y, z) is given by equation (1, 2, 3). Also show that 1 is the largest common factor for p and q and that they are not both odd numbers.
Assume (x,y,z) is a primitive Pythagorean triple, where x is odd, y is even and z is odd. Pythagorean triple. 1. x = p^2-q^2 2. y = 2pq 3. z = p^2 + q^2
k = p^2 and l = q^2. Check that the Pythagorean triple (x, y, z) is given by equation (1, 2, 3). Also show that 1 is the largest common factor for p and q and that they are not both odd numbers.
Assume (x,y,z) is a primitive Pythagorean triple, where x is odd, y is even and z is odd.
Explanation / Answer
(x,y,z) is aprimitive pythagorean triple.
x is odd, y is even and z is odd.
1. x = p2 - q2
2. y = 2pq
3. z = p2 + q2
x2 + y2 = (p2 - q2)2 + (2pq)2
= (p4 - 2p2q2 + q4) + 4p2q2
= p4 + 2p2q2 + q4
= (p2 + q2)2
= z2.
If p and q are both odd, then p2 and q2 are both odd too.
=> p2 + q2 and p2 - q2 are even.
=> x and z are even.
But z and z are both odd.
Therefore, we arrive at a contradiction and so p and q can't both be odd.
Let k be the gcd of p and q.
Let p = Pk and q = Qk
=> x = (Pk)2 - (Qk)2 = k2 (P2 - Q2)
y = 2PkQk = 2PQk2
z = (Pk)2 + (Qk)2 = k2 (P2 + Q2)
Then P2 - Q2, 2PQ and P2 + Q2 would also form a Pythagorean triple.
Since (x,y,z) is primitive, we arrive at a contradiction.
=> gcd(p,q) = 1.
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