Cfrgtkky

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

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

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

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 $b=2$

The general solution of $4b+3m=17$ is $b=17-3t$, $m=-17+4t$, with $tinmathbb Z$.

Now $b>0$ iff $tle 5$ and $m>0$ iff $tge 5$. Therefore, $t=5$ is the only solution. This gives $b=2$ and $m=3$.

hint

We have

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

thus

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

We have that

$$4b+3m=17 iff bequiv2 pmod 3quad land quad mequiv 1 pmod 2$$

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

• $k=1+r implies b=5+3r$


• $h=-1-2r implies m=-1-4r$

First $4cdot 4-3cdot 5=1$. So $(17cdot 4,-5cdot 17)=(68,-85)$ is a solution. All solutions are of the form $(68+kcdot 3,-85-kcdot 4)$.

The only way to get them both positive is if $k=-22$.

With, trivially $1le b le 4$ [because $4times 5 gt 17]$ and $1le m le 5 [6times 3 gt 17]$ 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.

• If $bgeq 9$ then (since $mgeq 1$) $$4b+3m>18+3m>17$$




• If $mgeq 6$ then (since $bgeq 1$) $$4b+3m>4b+18>17$$

Thus there are only finitely many solutions, $1leq b< 9$, $1leq m< 6$, and they can be checked directly.