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**hollywood** 1) Suppose n is not the square of an integer. Let n=ms^2, where m>1 has no square divisors. Let p be a prime divisor of m. We have z^2 = nd^2 = ms^2d^2, and since the powers of p in z^2, s^2, and d^2 are even, the power of p in m must be even. But then m has a square factor. By contradiction, therefore, n must be the square of an integer.

2) We need to prove that if d divides gcd(x,y) then d divides gcd(x,y,z) (forwards) and that if d divides gcd(x,y,z) then d divides gcd(x,y) (backwards). Then the set of divisors of gcd(x,y) will be the same as the set of divisors of gcd(x,y,z), so gcd(x,y)=gcd(x,y,z)

Forwards: if d divides gcd(x,y) then the previous proof shows that d divides z, so d is a common factor of gcd(x,y) and z, so d divides gcd(gcd(x,y),z) = gcd(x,y,z).

Backwards: if d divides gcd(x,y,z) = gcd(gcd(x,y),z), then it is a common factor of gcd(x,y) and z, and therefore divides gcd(x,y)