Thread: Perfect square

1) If gcd(x,3)=1 and gcd(y,3)=1, prove that cannot be a perfect square.

2) Given that gcd(a,b,c)lcm(a,b,c) = abc, prove that gcd(a,b)=gcd(b,c)=gcd(a,c)=1.

This is not homework, I'm doing it for extra practice, but I can't figure out how to prove these.

Any help would be appreciated!

[also under discussion in math link forum]

Originally Posted by kingwinner

1) If gcd(x,3)=1 and gcd(y,3)=1, prove that cannot be a perfect square.

Therefore .

So if , which is impossible.

Originally Posted by kingwinner

2) Given that gcd(a,b,c)lcm(a,b,c) = abc, prove that gcd(a,b)=gcd(b,c)=gcd(a,c)=1.

Fix a prime p, suppose the exponent of a,b,c of p are x,y,z respectively, WLOG suppose a<=b<=c. Show a=b=1

Originally Posted by FancyMouse

Fix a prime p, suppose the exponent of a,b,c of p are x,y,z respectively, WLOG suppose a<=b<=c. Show a=b=1

Q2) I've thought about it for awhile, but I can't figure out how to show a=b=1.

Also, aren't we supposed to show gcd(a,b)=gcd(b,c)=gcd(a,c)=1. How can we prove this?

Thanks!

Originally Posted by chiph588@

Therefore .

So if , which is impossible.

Thanks!
How about this?
"If x and y are odd, prove that cannot be a perfect square."

x odd => x=1(mod2) => = 1 (mod 2)
=> = 0 (mod 2)
=> ???

Originally Posted by kingwinner

Thanks!
How about this?
"If x and y are odd, prove that cannot be a perfect square."

x odd => x=1(mod2) => = 1 (mod 2)
=> = 0 (mod 2)
=> ???

is odd. For example, .

Originally Posted by chiph588@

is odd. For example, .

Sorry, maybe I wasn't clear. This is a different problem.
"If x and y are odd, prove that cannot be a perfect square."
How can we prove this?

Originally Posted by kingwinner

Sorry, maybe I wasn't clear. This is a different problem.
"If x and y are odd, prove that cannot be a perfect square."
How can we prove this?

Look at the definition of a Pythagorean triple (not necessarily primitive). One term is always a multiple of .

Originally Posted by chiph588@

Look at the definition of a Pythagorean triple (not necessarily primitive). One term is always a multiple of .

""If x and y are odd, prove that cannot be a perfect square."

How can we prove this directly by basic modular arithmetic?

x odd => x=1(mod2) => = 1 (mod 2)
=> = 0 (mod 2)
But I can't find a contradiction here...

Originally Posted by kingwinner

""If x and y are odd, prove that cannot be a perfect square."

How can we prove this directly by basic modular arithmetic?

x odd => x=1(mod2) => = 1 (mod 2)
=> = 0 (mod 2)
But I can't find a contradiction here...

This isn't a contradiction since is solvable.

Originally Posted by chiph588@

Look at the definition of a Pythagorean triple (not necessarily primitive). One term is always a multiple of .

So how can we prove this claim?
(this question is from the section on basic modular arithmetic; Pythagroean triple hasn't been introduced yet so I don't think we'll need it)

Thanks!

Originally Posted by kingwinner

So how can we prove this claim?
(this question is from the section on basic modular arithmetic; Pythagroean triple hasn't been introduced yet so I don't think we'll need it)

Thanks!

Why can't you just note that if then but the reduced residue class of is ?

Thanks! How to prove Q2?
2) Given that gcd(a,b,c)lcm(a,b,c) = abc, prove that gcd(a,b)=gcd(b,c)=gcd(a,c)=1.

Originally Posted by kingwinner

Q2) I've thought about it for awhile, but I can't figure out how to show a=b=1.

Also, aren't we supposed to show gcd(a,b)=gcd(b,c)=gcd(a,c)=1. How can we prove this?

Thanks!

Sorry there's a typo. I meant x=y=1, not a=b=1. i.e. if p^x || a (i.e. p^x divides a but p^{x+1} doesn't divide a), p^y || b, p^z || c, and x<=y<=z, then x=y=1, which implies that if some prime one of a,b,c is divisible by p, then only one of them is divisible by p