Show that $\gcd(5a+8, 7a+3)=1\textrm{ or } 41$. Best way to solve a problem of this sort?

Question

Let $a\in \mathbb Z$. Show that $\gcd(5a+8, 7a+3)=1\textrm{ or } 41$. Knowing that $\forall k \in\mathbb Z, \gcd(a,b)=\gcd(b, a-kb)$.

I actually know how to solve this particular problem: $\gcd(5a+8, 7a+3)=\gcd(5a+8,2a-5)=\gcd(a+18,2a-5)=\gcd(a+18:-41)$. So $(a+8:-41)\in \{1, 41\}$.

What I want to know is if there's an efficient way to solve any problem of this sort. The goal being to eliminate variables from either side of the gcd no matter the parity of the terms involved. Manually adding and substracting to each sum to have the variable $a$ cancelled. This is only the simplest example. Another one is: $\gcd(2a^2+3a-1, 5a+6)$.

Any ideas?

You're just going through the Euclidean algorithm except with expressions instead of integers. — , May 14 '21 at 19:03
The standard methods are all described in the linked dupes (and their links). — Bill Dubuque, May 14 '21 at 20:19

score 0 · Answer 1 · answered May 14 '21 at 18:59

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Say there is $d$ which divide both. Then $d\mid 7(5a+8)$ and $d\mid 5(7a+3)$ so $d$ divide their difference: $$d\mid 7(5a+8)-5(7a+3) = 41$$

So $d\in\{1,41\}$.

answered May 14 '21 at 18:59

nonuser

90,026

Please strive not to add more dupe answers to dupes of FAQs, cf. site policy here. – Bill Dubuque May 14 '21 at 19:18
But my ansewr is totally of different kind. @BillDubuque – nonuser May 14 '21 at 19:37
Not true. It is the exactly the same as the first method in this answer in the first linked dupe (and is presented here many times, as anyone experienced in ENT here would surely know). Be sure to read the linked page on recent site policy changes which pertains to matters like this. – Bill Dubuque May 14 '21 at 20:14
How on earth could I find that post of yours from 6 years ago? @BillDubuque – nonuser May 16 '21 at 13:39

Show that $\gcd(5a+8, 7a+3)=1\textrm{ or } 41$. Best way to solve a problem of this sort?

1 Answers1