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Let us have two numbers, a, b, so that gcd(a,b)=d.

Let us have two prime numbers k,l so that gcd(ak, bl)=d.

Is this valid for all prime numbers or is there any conditions that must be met?

I guess that it is valid for all prime numbers except those that are divisors of a and b, i.e. kcannot be the prime number which is also divisor of b and lcannot be the prime number which is also divisor of a, but I do not know how to prove it.

Could anyone help? Thank you.

Bill Dubuque
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ZuzanaTelefony
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  • Try some examples. Maybe $a=6$ and $b=10$. What can the prime numbers $k$ and $l$ be to get $gcd(6k,10l)=2$? – ndhanson3 May 21 '21 at 07:47
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    Are you asking whether $\gcd(a,b)=d$ implies $\gcd(ak,bl)=d$ for all $a,b,k,l\in\mathbb{Z}$ with $k$ and $l$ primes? – Ellefee May 21 '21 at 07:50
  • @ndhanson3, I tried the examples, but I just cannot figure out the mathematical proof for that – ZuzanaTelefony May 21 '21 at 07:50
  • @Ellefee, I am asking what is the mathematical proof for the conditions that must be met so that gcd(ak,bl)=d for when gcd(a,b)=d. – ZuzanaTelefony May 21 '21 at 07:51
  • @ZuzanaTelefony What examples have you tried? Have you found any patterns?

    You need to have an idea of what the conditions are before trying to prove anything, so let's back up and see what they might be.

     – ndhanson3 May 21 '21 at 07:52
  • @ndhanson3, e.g. for your particular a=6 and b=10 ... k and l can be practically every prime number the prime numbers included already in a and b. (e.g. k=3 and l=7 or k=31 and l=37. BUT NOT k=5 and l=3) – ZuzanaTelefony May 21 '21 at 07:58
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    @ZuzanaTelefony Add your conjecture/hypothesis/guess to the question! – ndhanson3 May 21 '21 at 08:00
  • Cancelling $,d,$ yields $,\gcd(ak,bl) = d !\iff! \gcd(\bar ak,\bar bl) = 1,$ for $,\bar a = a/d,, \bar b = b/d,,$ then apply the linked dupe using $,\gcd(\bar a, \bar b) = 1 = \gcd( k,l)\ \ $ – Bill Dubuque May 21 '21 at 08:06

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