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I am having a lot of trouble with a GCD problem. The problem is find $\gcd(5a+3,4a+7)$ for $a>0.$ I have been working on this for quite a while now and have the following issues:

I first tried to apply the euclidean algorithm, but the problem was that I had no idea how to deal with a when finding r. It didn't make much sense to try and divide two polynomials in terms of a.

So I began trying a few values of a.

After trying a few values, I thought they may be co-prime. so I tried proving this by showing x(5a+3) + y(4a+7) = 1. This implies that a(5x+4y)+(3x+7y) = 1, and so 5x+4y =0 and 3x+7y = 1. This doens't have a solution in the integers.

So I began trying to prove this by induction. This didn't work either. Specifically, I had no idea of showing if d|(5a+8) then d|(5a+3) (similarly for the other value) to derive a contradiction.

So I tried more values. After checking the first 2000 digits, I realized 1 and 23 are the only possible values for the gcd (atleast up to 2000). Specifically, whenever a has the form 27+23j, we get that they have 23 as a common factor. Similarly, when a=4 we get the case where they both equal 23.

Otherwise, they are reletively prime.

I can prove that if a = 27+23j, then their gcd is 23 using the euclidean algorithm:

5(27+23j) +3 = [4(27+23j)+7] + 23+23j

4(27+23j)+7 = (23+23j)4 + 23

(23+23j)4 = 23(4+4j) + 0

And so gcd(5(27+23j)+3,4(27+23j)+7) = gcd(23(4j+4),23) = 23 for any j.

But how do I prove that if a is not 4, and is not of the form 27+23j for some j, then gcd(5a+3,4a+7)=1? I am not sure where to proceed with this proof, how to contrain a to make it more wieldy.

The difference in this question from the similar one posted before is twofold: firstly, the polynomials we’re considering are different. Second, my problem is that of proving that d=1 when certain conditions hold for a, not that of finding d given certain conditions for a. I’m confident my conjecture is correct, but am in need of techniques to help me prove it.

Mani
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    $5(4a+7)-4(5x+3)=23.$ So the GCD can be what values? – Thomas Andrews Jun 30 '23 at 13:54
  • Compute $\operatorname{gcd}(5a+3,4a+7)$ for $a=0,1,\dots$ and get $1, 1, 1, 1, 23, 1, 1, 1, 1, 1, 1,\dots$. Then use this to conjecture an answer. – GEdgar Jun 30 '23 at 14:02
  • Hint: $\bmod d!:\ 5a+3\equiv 0\equiv 4a+7,,$ so eliminating $,a,$ yields $\ldots$ See the linked dupe for this and other methods. – Bill Dubuque Jun 30 '23 at 14:06
  • I’m still a bit confused about this. The issue is, when I mod by d, I get that 5a+3 = 4a+7 and so a = 4 mod d. Thus d|a-4 (this is where 27 comes in I imagine). But I’m not sure how this proves the cases where d must be = 1. So if a is not 4 or a member of the progression 27+23j, then still a-4 can be anything. It can be that a=294, and so d|294. I’m not sure how this proves that d=1 in this scenario. – Mani Jun 30 '23 at 14:20

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