Let $p$ and $q$ be distinct primes. Prove that $pq$ is not expressible in the form $px+qy$ where $0 \leq x \leq q-1$ and $0 \leq y \leq p-1$. Similarly prove that $pq-p-q$ is not expressible in that form.
I was thinking of proving this by proof by contradiction, but I couldn't think of a nice way doing that.
Assuming the primes are different:
$$pq=px+qy\implies p(q-x)=qy\implies p\,\mid\,y\;,\;\;\text{and likewise}\;\;q\,\mid\,x\implies$$
$$pq=pqx'+pqy'\implies x'+y'=1$$
Try now to complete the argument taking into account your question´s conditions.
By contradiction, assume that $pq=px+qy$ for distinct primes $p,q$ and some $x$ and $y$ as in the statement of the question. If you reduce this equation modulo $p$ you get $$qy=0\mod p.$$ As $p$ and $q$ are (particularly) coprimes this implies that $y=0\mod p$, so by the bounds on $y$ you get $y=0$. Similarly reducing the equation modulo $q$ you get $x=0$ so you have a contradiction.
The same technique gives you de desired conclussion for $pq-p-q$.
If $p$ and $q$ are same then we can easily find a solution.
Now let $p,q$ be different. Obviously $x,y >0$, as otherwise LHS is always less than RHS. Now assume that there exist $x,y \in \mathbb{N}$. Then working modulo $p$ we get that $p \mid y$, but as $1 \le y \le p-1$ we obtain a contradiction.
I assume that $p$ and $q$ are distinct.
If there exists $(x,y)\in\{0,\cdots,q-1\}\times\{0,\cdots, p-1\}$ such that: $$pq=px+qy.$$ Then, $qy$ is divisible by $p$, but since $p$ and $q$ are coprime, $y$ is divisible by $p$. Therefore, $y=0$. In the same way, $x=0$. Whence, $pq=0$, which is not the case.
A slight generalization yields insight. By the Lemma, all solutions of $\, px + q y = pq\,$ arise from the particular solution $\,(x,y) = (q,0)\,$ by adding or subtracting multiples of $\,(q,-p).\,$ So the only solution with $\,0\le x < q\,$ is $\,(q,0) - (q,-p) = (0,p),\,$ which fails $\,0\le y < p$
Lemma $\ pX\!+\!qY = px\!+\!qy\ \Rightarrow\ (X,Y) = (x,y) + k(q,-p)\,$ for some integer $\,k$
Proof $\,\ p(X\!-\!x) = q(y\!-\!Y)$ so $\,p,q\,$ coprme $\,\Rightarrow\,p\mid y\!-Y,\,$ so $\,y\!-\!Y = kp,\,$ so back substituing $\, X\! -\! x = kq,\,$ so $\,(X,Y)-(x,y) = (X\!-\!x,Y\!-\!y) = (kq,-kp) = k(q,-p)$
Remark $\ $ If you've studied linear algebra or differential or difference equations (recurrences) then you may recognize this is a special case of the basic fact that the general solution of a nonhomogenous linear equation is given by any particular solution plus the sum of the general solution of the associated nonhomogeneous equation.
Here the homogenous form is $\,px + qy = 0\,$ whose nonzero solutions satisfy $\,y/x = -p/q.\,$ So the nonzero solutions are the fractions equivalent to $-p/q$, which are of the form $\,-kp/kq,\,$ i.e. $\,(x,y) = (kq,-kp) = k(q,-p)$ as above. That all equivalent fractions have this form depends on $\,p,q\,$ being coprime, i.e. the fraction is in lowest terms - see unique fractionization for more.
