# Has this basic number theory conjecture been discovered or proved before?

So I was sitting in a car bored for two hours playing around with the calculator on my phone, and I discovered that by picking two numbers (at least one being odd and both being not divisible by each other) I could always add them together to make many different numbers. Furthermore, I could always add and subtract them together make 1; this is significant because I could then negate and indefinitely duplicate the series to create any positive and negative integers. Here are some examples:

8-7=1
16+16+16+16-9-9-9-9-9-9-9=1
17+17+17-5-5-5-5-5-5-5-5-5-5=1


and so on.

For convenience, I preferred to write these equations in the form a*x-b*z=1

 8*1-7*1   = 1
16*4-9*7   = 1
17*3-5*10  = 1


Here is my Conjecture:

Any two non-zero integers indivisible of each other, if at least one of them is odd, can be added or subtracted together indefinitely to make any integer.

Another way of writing it (though it is more limiting) is this:

Given that a and b are two non-zero integers indivisible of each other, if at least one of them is odd, there always exists some value of x and z for any given integer n that makes the equation ax-bz=n true.

My question is, has anyone ever discovered or proved this conjecture before, and how can it be proven? It almost feels like the solution should be as obvious as the equation a*x-b*z=0 (which just requires finding the lcm of the two numbers) and that it's just a simple quirk in numbers based upon the already laid down basic laws of mathematics.

• The integers obtainable from integral linear combinations of $a$ and $b$ are precisely the multiples of $\gcd(a,b),\$ i.e. $\ a\Bbb Z+b\Bbb Z = \gcd(a,b)\Bbb Z,\,$ see this question – Bill Dubuque Oct 28 '19 at 3:36
• 1st paragraph says "at least one being even", conjecture says "at least one of them is odd". Note that if they are coprime then it is guaranteed that at least one is odd. Note also that even if both $a,b$ are odd you can still get any $n$ as $ax-by$ (provided of course $a,b$ are relatively prime). – Gerry Myerson Oct 28 '19 at 4:55
• @GerryMyerson Gotcha. I replaced "even" with "odd" as I meant odd. I also reverted "co-prime" to indivisible as it fits more with the answer below and with your comment. – Zach K Oct 29 '19 at 21:00
• Coprime is what you want. The answer by Don shows that even if neither one of $a,b$ divides the other one, you may not be able to get $1$; to get that, the necessary and sufficient condition is that $a,b$ be coprime. – Gerry Myerson Oct 29 '19 at 21:21
• @GerryMyerson I understand that co-prime is what I want, and that was the original reason I modified my post to include it; however, my original post contained "indivisible" instead of "co-prime", and Don responded to my post while it was in that state (which is why he said my conjecture was false). I didn't modify my answer out of ignorance, I put it back to its original state. – Zach K Oct 30 '19 at 4:29

This is false: Try $$15$$ and $$21$$. Note that $$3\mid15x-21y$$ for all $$x,y$$.

However, the rule you are thinking of is basically Bezout's Identity.

• God, why do all of my low-effort answers get upvoted, and the ones I spend hours on get ignored :( – Don Thousand Oct 28 '19 at 3:24
• My bad, I should've said co-prime, not indivisible. I will accept your answer when I can. – Zach K Oct 28 '19 at 3:24
• @DonThousand well, in 2 lines you answered both what OP asked and what OP wanted to ask so... you have my +1 – Ripstein Oct 28 '19 at 11:34
• because most people don't know what you mean by the hours long answers ? – user645636 Oct 28 '19 at 12:03