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A previous professor I had last year sent me this question, which he says he found an answer but it's too long, and asked me to try and solve it; though after solving it, my answer was a little long, so I'm curious to see if there's any shorter solution. The problem is the following:

Given $p,q$ positive coprime numbers, prove that there exists $a$, $b$ positive integers such that $ap+bq=pq-1$.

I began seeing that by Bezout's identity there exist $A,B\in\mathbb{Z}$ such that $Ap+Bq=1$, then multiplying that by $pq-1$ you get the following: $$A(pq-1)p+B(pq-1)q=pq-1$$ But we don't know if $A(pq-1)$ and $B(pq-1)$ are positive, so what I managed to do was to try to find positive solutions starting from those. What I did was set a general solution (of the equation $a'p+b'q=pq-1$) using that initial solution, and try to find a certain parameter $k$ which could make the solutions, let's call them $A_k$ and $B_k$ both positive; I hope this is enough explanation, otherwise I can try to explain more what I did but it would take me quite some time.

mrod1605
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    We can handle a page and a half, let's see. – David P Mar 08 '22 at 20:53
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    Welcome to Math.SE! ... If you don't include your solution, no one can tell if their solution is different. This increases the possibility that someone will waste time duplicating your effort or explaining concepts you already understand. – Blue Mar 08 '22 at 20:55
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    Hint: don't scale $,Ap+Bq=1,,$ instead subtract it from $,pq.\ \ $ – Bill Dubuque Mar 08 '22 at 23:13
  • Yeah, I should have seen that. Either way my solution is quite an overkill lol, I still like it tho. – mrod1605 Mar 10 '22 at 11:14
  • This follows immediately from basic results on the Frobenius coin problem, e.g. by here in the dupe, $,px + qy = n > pq-p-q,$ always has solutions $,x,y\ge 0.,$ OP is special case $,n = pq-1,$ (note $,x\neq 0,$ else: $\ qy = pq-1\Rightarrow q\mid 1;,$ similarly $,y\neq 0)$ – Bill Dubuque Mar 10 '22 at 18:13

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I am going to demand $p,q \geq 2.$ Treat one or both as $1$ as special cases for later consideration.

Begin with $A_0p+B_0q = 1.$ Take $A = A_0 + s q$ so that $ 0 \leq A < q.$ Note, with $q \geq 2,$ we actually have $0 < A < q.$ Then we have $B = B_0 - s p,$ we do have $Ap+Bq=1,$ we know $A,B \neq 0.$ Bounds on $B$ come from $ 0 < Bq+Ap < Bq+qp = (B+p) q.$ That is, $p+B > 0.$ And $$ 0 < A < q \; \; \; , \; \; \; -p < B < 0 $$ But, you see, $$ (q-A)p + (-B)q = pq-1 $$

Will Jagy
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