Prove that for any primitive Pythagorean triple x, y, and z, each of the following hold.
a) 3 divides x or 3 divides y
b) 5 divides x or 5 divides y or 5 divides z
c) 12 divides xy
d) 60 divides xyz
Answers
Answer:
By primitive it is meant that x, y and z have no common factor greater than 1.
a) Dividing a number by 3, the possible remainders are 0, 1 and 2.
So square numbers leave remainders of 0 or 1 (since 2²=4, which leaves a remainder of 1).
In terms of modular arithmetic, the quadratic residues modulo 3 are 0 and 1.
If neither x nor y were divisible by 3, it follows then that x² ≡ y² ≡ 1 (mod 3); that is, both x² and y² leave a remainder of 1 when divided by 3. Then z² = x² + y² ≡ 1 + 1 = 2 (mod 3), so z² leaves a remainder of 2 when divided by 3. But we've already seen that squares always leave a remainder of 0 or 1, so this is not possible. It follows that 3 divides x or 3 divides y after all.
For the following, it'll be shorter to use the language of modular arithmetic more. If you're not very familiar with this, compare with the above paragraph to make sense of things.
b) Of course, each of x, y and z is congruent to one of 0, 1, 2, 3 or 4 modulo 5. So x², y² and z² are each congruent to one of 0, 1, 4 modulo 5. [ Notice that 3² = 9 ≡ 4 (mod 5) and 4² ≡ (-1)² = 1 (mod 5). ]
Suppose neither x nor y is divisible by 5.
If x²≡y²≡1 (mod 5), then z²=x²+y²≡1+1=2 (mod 5), which is not possible.
If x²≡y²≡4 (mod 5), then z²=x²+y²≡4+4≡3 (mod 5), which is not possible.
If {x²,y²} ≡ {1,4} (mod 5), then z²=x²+y²≡1+4=0 (mod 5), so z is divisible by 5.
It follows that 5 must divide one of x, y or z.
c) Each of x², y² and z² is congruent to either 0, 1 or 4 modulo 8. [ Notice that 3² = 9 ≡ 1 (mod 8), etc. ]
If x²≡y²≡1 (mod 8), then z² = x²+y² ≡ 1+1 = 2 (mod 8), which is not possible.
We cannot have x²≡y²≡4 (mod 8) as that would mean that x and y are both even, contrary to (x,y,z) being a primitive Pythagorean triple.
If {x²,y²} ≡ {1,4} (mod 8), then z²=x²+y²≡1+4=5 (mod 8), which is not possible.
It follows that one of x² or y² must be congruent to 0 (mod 8), and therefore, either x or y is a multiple of 4.
Together with the result from (a), it follows that 12 divides xy.
d) As 12 divides xy by (c), and 5 divides xyz by (b), it follows that xyz is divisible by 60.