0

This is a sub-problem that I am solving in light of a larger more general problem. I believe it is probably quite easy but I often struggle with number theory. Here is my theorem: the number $m$ is given as $$m = (p-1)(q-1)-1 = p(q-1) - q$$

What I can prove (I think) is that this is an upper bound for $m$. Here goes:

Suppose that $m$ is actually some larger number $$\tilde{m} = p(q-1) - q + c$$ Then, as $\gcd(p,q) = 1$, by Bezout, there exist $a,b$ such that $c= ap+bq$ with $b >0$ and $|a| < \frac{q}{\gcd{p,q}}$. Now, $$\tilde{m} = p (q- 1 +a) + (b-1)q. $$ Now by the bounds on $a$ and $b$ we have that $b-1 \geq 0$ and $q -1 + a \geq 0$, thereby $\tilde{m}$ does not meet the desired requirements.

So that is my proof so far, I believe it is correct. What remains is to show that the chosen $m = p(q-1) - q$ actually cannot be written as $ap+bq $ for $a,b \geq 0$. Off course in the representation I choose the coefficient for $q$ is $-1$ but I can't really see how I can rule out other representations for which the coefficients are non-negative. Thanks in advance for any help!

Slugger
  • 5,556

1 Answers1

1

Ad absurdum:

Let $a,b>0$ be s.t. $m=ap+bq$

Since $m<(p-1)(q-1)$ we also have $a<q-1$ and $b<p-1$.

Edit: to prove this, notice that if $a\geq q-1$ then $ap+bq>ap\geq (q-1)p>(p-1)(q-1)>m$, which contradicts $m=ap+bq$.

Then $m=p(q-1)-q=ap+bq$, hence $(q-1-a)p=(b+1)q$, where $0<q-1-a<q$ and $0<b+1\leq p$.

This contradicts the fact that $p$ and $q$ are coprime.

Evargalo
  • 2,593