Multiples of 2 numbers that differ by 1

Asked Jan 03 '20 at 16:35

Active Jan 03 '20 at 20:44

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I have 2 known positive integers, $a$ and $b$ . Is there a ' standard ' formula to find the lowest (if possible) positive integers $x$ and $y$ , so that the following is true?

$$xa = yb + 1$$

edited Jan 03 '20 at 16:39

The Demonix _ Hermit

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asked Jan 03 '20 at 16:35

DanDeg

1

Have you seen Bezout's Lemma? – Rushabh Mehta Jan 03 '20 at 16:37
Are you familiar with modular arithmetic and congruency ? – The Demonix _ Hermit Jan 03 '20 at 16:37
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Look up the 'Euclidean algorithm' – Empy2 Jan 03 '20 at 16:37
Take $a=2$ and $b=4$. Then there are no such integers $x,y$, because $xy-yb$ is even. For coprime $a$ and $b$ just use Bezout's Lemma. – Dietrich Burde Jan 03 '20 at 16:37
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Integers $x,y$ can be found iff $\gcd(a,b)=1$. Since in at least some cases one of $x,y$ may be negative, it is not clear what you mean by making $x,y$ the lowest. One possibility is $|xy|$ is minimized. – Keith Backman Jan 03 '20 at 16:51
I only know that b will always be prime. And it doesn't necessarily need to be the 'lowest' answer, I just need a answer. Thanks for the response by the way! – DanDeg Jan 03 '20 at 18:54
@DanDeg So what you seek is the inverse of $,a,$ modulo a prime $,p,,$ for which there are various well known algorithms - see the linked dupes. – Bill Dubuque Jan 03 '20 at 20:43

Multiples of 2 numbers that differ by 1

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