How can you prove that $b=2, m=3$ are the only positive integer solutions to $4b+3m=17$?

Question

How can you prove that $b=2, m=3$ are the only positive integer solutions to $4b+3m=17$ without a proof by exhaustion?

Answer 1

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

• $k=1+r \implies b=5+3r$
• $h=-1-2r \implies m=-1-4r$

Answer 2

• 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.

Answer 3

hint

We have

$$4\times 4-3\times 5=1$$ $$4(b-4)+3m=1$$

thus

$$4(8-b)=3(m+5)$$

Answer 4

$$ 4b+3m=17 \implies 4b=17-3m $$

$$\implies 4b=16+(m+1)-4m = 4(4-m)+ (m+1) $$

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 $b=2$

Answer 5

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 $k=-22$.

Answer 6

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.

Answer 7

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 $m=3$.