Question

Let a and b be square numbers. Is gcd(a,b) also square? Prove your answer, or find a counterexample. (Hint: First show that a number is square if and only if its prime decomposition contains only even exponents.)
_i solved the hint~ but have no idea of next step~

Answer 1

1,4,9,16,25 etc.... these are square numbers
because
1^2 = 1, 2^2 = 4, 3^2 = 9 4^2 = 16 and 5^2 = 25
all with even exponents

Answer 2

100 and 4. gcd is 2, 4 and 9. gcd is 1. A lot of other examples to prove..gcd of two square numbers is not always a square number

Answer 3

dont google that~ the statement is actually true(can be proved), btw gcd(100,4)=4.

Answer 4

another hint : square root of gcd exists in both the numbers :)

Answer 5

yea that makes sense, but i dont know how. omg, so tricky '

Answer 6

Okay. heres the proof :

since a & b are square numbers, they can be written as even powers of prime nuumbers : p,q,r,t

let :
\(\huge a = p^{2}q^{8}r^{6} \\ \huge b = p^2t^4r^4\)

Answer 7

@satellite73 :)

Answer 8

gcd is a square because gcd itself can be factored as the product of primes
say \(p_1^{\alpha_1}\times p_2^{\alpha_2}\times ...\times p_n^{\alpha_n}\)
but if \(p_i|a\) then so does \(p_i^{2k}\) for some \(k\) since you have proved that all the
exponents are even
similarly \(p_i^{2j}\) divides \(b\) and since \(p_i^{\alpha_i}\) divides both \(a\) and \(b\) for each \(i\) you have \(\alpha_i=2l\) for some \(l\)

Answer 9

thx u people for the hints~ helps a lot. i think i got the idea :D

Answer 10

I wonder if we can prove this without referring to prime factorization. It's true that

\[\gcd(a^{2},b^{2}) = (\gcd(a,b))^2\]

but how do you prove it? Let's write

a = c * gcd(a,b)
b = d * gcd(a,b)

with gcd(c,d) = 1

so we have

\[a^{2} = c^{2} * (gcd(a,b)^{2}\]
\[b^{2} = d^{2} * (gcd(a,b)^{2}\]
\[gcd(c^{2},d^{2}) = 1\]

So on the one hand, we have

\[(gcd(a,b))^{2} | a \]
\[(gcd(a,b))^{2} | b \]

proving that it's a common divisor. On the other hand if k is a divisor of a, b and \(gcd(a,b)^2\) it is of the form \(k*gcd(a,b)^2\) where k divides both \(c^2\) and \(d^2\). So k =1 proving that

\[\gcd(a^{2},b^{2}) = (\gcd(a,b))^2\]

Answer 11

wow~ thank you for this~ :) @beginnersmind