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Following the next theorem:

$$\gcd(a; b) = \gcd(a - b; b)$$

I get to the point where from these two numbers $4n + 3$ and $20n + 23$ I get these: $4n + 3$ and $8$.
It seems obvious that these numbers don't have any common divisor except $1$, but I don't recall any applicable axiom/theorem that would help me to prove that. So my question may seem obvious, but how to prove it?

Robert Z
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curioushuman
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    $4n+3$ is always odd. While $8 = 2^3$ – Nikunj Mar 22 '23 at 09:03
  • I've been thinking about it and I don't understand. How does it prove it? Okay, the first number is always odd, the second one is even + a power of 2. But, for instance, pairs of numbers such as (49; 14), (243; 36) are odd and even relatively, but they are not coprime. The first pair of numbers both can be divided by 7, and the second one - by 9. I understand it has to do something with the fact that 8 is 2^3, but what exactly? Because I don't understand how it proves. – curioushuman Mar 22 '23 at 09:33
  • If $c$ is coprime to $d$ then so too is $c^3,,$ by coprimes to $d$ are closed under products, by the linked dupe. The OP is special case $,c=2,, d = 4n+3.,$ Please delete the question once all is clear - we already have too many cases of this. – Bill Dubuque Mar 22 '23 at 09:41
  • The divisors of 8 are 1, 2, 4, 8. The gcd divides 8 so it must be one of those. The gcd must also divide 4n+3, an odd number, i.e. a number that does not have a factor 2 in it. So..... – Jaap Scherphuis Mar 22 '23 at 09:42
  • The only prime factor of $2^k$ is $2,$ and $2$ is not a prime factor of $4n+3.$ Therefore $gcd(2^k, 4n+3)$ must be $1.$ – Adam Rubinson Mar 22 '23 at 10:23

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