How can you prove that $b=2, m=3$ are the only positive integer solutions to $4b+3m=17$? How can you prove that $b=2, m=3$ are the only positive integer solutions to $4b+3m=17$ without a proof by exhaustion?
 A: $$ 4b+3m=17 \implies 4b=17-3m $$
$$\implies 4b=16+(m+1)-4m = 4(4-m)+ (m+1) $$
Thus $m+1$ must be a multiple of $4$ where $m<4$ to make  both sides positive.
The only choice is $m=3$ which implies $b=2$ 
A: The general solution of $4b+3m=17$ is $b=17-3t$, $m=-17+4t$, with $t\in\mathbb Z$.
Now $b>0$ iff $t\le 5$ and $m>0$ iff $t\ge 5$. Therefore, $t=5$ is the only solution. This gives $b=2$ and $m=3$.
A: We have that
$$4b+3m=17 \iff b\equiv2 \pmod 3\quad \land \quad m\equiv 1 \pmod 2$$
that is


*

*$b=2+3k$

*$m=1+2h$
and
$$4b+3m=17 \iff 4(2+3k)+3(1+2h)=17 \iff 12k+6h=6 \iff 2k+h=1$$
that is by Bezout's identity


*

*$k=1+r \implies b=5+3r$

*$h=-1-2r \implies m=-1-4r$
A: hint
We have
$$4\times 4-3\times 5=1$$
$$4(b-4)+3m=1$$
thus
$$4(8-b)=3(m+5)$$
A: First $4\cdot 4-3\cdot 5=1$.  So $(17\cdot 4,-5\cdot 17)=(68,-85)$ is a solution.   All solutions are of the form $(68+k\cdot 3,-85-k\cdot 4)$.
The only way to get them both positive is if $k=-22$.
A: With, trivially $1\le b \le 4$ [because $4\times 5 \gt 17]$ and $1\le m \le 5 [6\times 3 \gt 17]$ you don't have many cases to check. But you can work more efficiently.
Note that because $b$ and $m$ are both at least one (positive integers) you can reduce this to $4c+3n=10$ for non-negative $c$ and $n$ (may now be zero) with $b=c+1$ and $m=n+1$.
Now $10$ is not a multiple of $4$ or a multiple of $3$ so you can't have either $c=0$ or $n=0$ so both must be at least $1$ whence $4d+3p=3$ with $b=d+2$ and $p=m+2$ and $d$ and $p$ are non-negative.
Next $d$ can't be positive so must be zero. and $p=1$ is the only possibility.
A: *

*If $b\geq 9$ then (since $m\geq 1$) $$4b+3m>18+3m>17$$

*If $m\geq 6$ then (since $b\geq 1$) $$4b+3m>4b+18>17$$
Thus there are only finitely many solutions, $1\leq b< 9$, $1\leq m< 6$, and they can be checked directly.
