Is this GCD statement true?

Question

If we have integers $h$, $i$, $j$, and $k$, would it be true to say that $\gcd(h,i)\gcd(j,k)|\gcd(hi,jk)$? If so, how can we prove it?

Maybe you want $\gcd(h,j)\gcd(i,k)|\gcd(hi,jk)$? – Ross Millikan Oct 14 '11 at 02:56 — Ross Millikan, Oct 14 '11 at 02:56

score 5 · Accepted Answer · answered Oct 14 '11 at 02:53

5

No, consider $h=i=2$ and $j=k=3$, then $$\gcd(h,i)\gcd(j,k)=\gcd(2,2)\gcd(3,3)=2\cdot 3=6\nmid 1=\gcd(4,9).$$

answered Oct 14 '11 at 02:53

Zev Chonoles

129,973

score 2 · Answer 2 · answered Oct 14 '11 at 03:14

2

By distributivity $\rm\ (h,i)\:(j,k) = (hj,hk,ij,ik)\ $ which divides $\rm\ hj, ik\ $ so also $\rm\:(hj,ik)\:.\:$ Presumably that's what was intended.

answered Oct 14 '11 at 03:14

Bill Dubuque

272,048

score -2 · Answer 3 · edited Jun 12 '20 at 10:38

-2

This statement is false, You will see it whenever you take $h,i,j,k$ such that the products $hi$ and $jk$ are coprime but the numbers $h,i$ and $j,k$ are not pairwise coprime.

Illustrations:

Select $h=2, i=4$ and $j=3,k=9$, you will see that $6\not| 1$.

Select $h=5,i=15$ and $j=2,k=4$, you will again see that $10\not|1$.

edited Jun 12 '20 at 10:38

Community

1

answered Dec 22 '16 at 18:20

Vidyanshu Mishra

10,143

4

What was the point of answering a 5 year old question? – Jyrki Lahtonen Dec 22 '16 at 18:31
@JyrkiLahtonen As here was nothing for me to answer in today's asked questions and Also this question needed a good answer (Which I think mine is...).. – Vidyanshu Mishra Dec 22 '16 at 18:33
Also @JyrkiLahtonen, It was not for reputations as I have already got the daily cap 3-4 hrs ago I think. – Vidyanshu Mishra Dec 22 '16 at 18:33
Bhai, tere number pe call nhi lag raha hai, time mile to call back kariyo... KVPY ka result aa gaya hai. – Jaideep Khare Dec 22 '17 at 17:29

Is this GCD statement true?

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