prove that if x square + y square = z square then one of x,y is a multiple of 3 and one of x,y,z is a multiple of 5
Answers
Step-by-step explanation:
If x2+y2=z2, how do I prove that xyz is a multiple of 60?
We will restrict ourselves to pythagorean triplets that are coprime. If not, we can reduce the original triplet dividing each number by the g.c.d. of the three, and if this new triplet satisfies the condition that its product is a multiple of 60, the original one satisfy it too.
The possible remainders when a number is divided by 4 are 0, 1, 2, and 3. If the number is a perfect square, the possible remainders are 0 and 1. If x and y are both odd, the remainders of x2 and y2 would be both 1, and therefore the remainder of z2 would be 2, which is impossible. Therefore, at least one from x and y must be even. If both are even, so would be z, and therefore the triplet would not be coprime. Then, we will assume that x is even and y is odd, and in this case z must be odd too. Let’s assume that x is not a multiple of 4. We have
x=4k+2
y=2p+1
z=2q+1,
and hence
16k2+16k+4+4p2+4p+1=4q2+4q+1
which can be written as
16k2+16k+4+4p(p+1)=4q(q+1).
Both p(p+1) and q(q+1) are even numbers, therefore can be replaced by
16k2+16k+4+4(2m)=4(2n).
or
8(2k2+2k+8m)+4=8n,
but this is impossible. Therefore, x must be a multiple of 4, and so is the product xyz.
The possible remainders when a number is divided by 3 are 0, 1, and 2. If the number is a perfect square, therefore, the possible remainders are only 0 and 1. We can easily see that both x2 and y2 cannot have a remainder 1, because z2 cannot have a remainder 2. Then, from x and y at least one must be a multiple of 3. Therefore, the product xyz is divisible by 3.
The possible remainders when a number is divided by 5 are 0, 1, 2, 3, 4, and 5. If the number is a perfect square, the possible remainders are 0, 1, and 4. If neither x nor y are multiples of 5, the possible remainders of x2 and y2 are 1 and 4. If they are both 1, the remainder of z2 would be 2, which is impossible. If they are both 4, the remainder of z2 would by 3, which is impossible. Then, the remainders must be 1 and 4, and the remainder of z2 is 0. In any case, the product xyz is a multiple of 5.
We’ve found that xyz is a multiple of 4, 3, and 5. Therefore, it must be a multiple of 60 .