are there natural numbers $a,b$ such that $b=\frac{(a-2)}{(a-4)}$? (with $a>6$)

Question

by just trying setting $a$ equal to a bunch of different (even) natural numbers, i'm pretty sure there are no such $a,b$ - but i'm stuck trying to prove it. i was thinking a proof by contradiction: if such $a,b$ exist then $(a-2)=(a-4)c$, where $c$ is some natural number, and then $b=4c$ (any multiple of 4??) which definitely isn't true. but again, i don't really know how to go about proving that... any help with this would be greatly appreciated!

Answer 1

If $a>6$, then $a-4>2$, so $0<\frac{2}{a-4}<1$. Note that $b=\frac{a-2}{a-4}=1+\frac{2}{a-4}$, so there are no such natural numbers $a$ and $b$.

Answer 2

No proper divisor of $n$ can be greater than $\frac n2$, so your assumption implies that $a-4≤\frac a2-1$ which implies $a<6$.

Answer 3

You can rewrite this as $$ab-4b-a+2=0$$ or $$(a-4)(b-1)=2$$

Now $b\gt 0$ so the only options for $b-1$ to be integers are $b=2,3$ with corresponding values for $a$ as $6,5$.

Rearranging again, if you want a more analytical proof gives $$a=4+\frac {2}{b-1}\le 6$$

and with $b-1\ge 1$ we have $2\le 2(b-1)$ and $\frac {2}{b-1}\le 2$ as required.

Answer 4

If $a-4$ divides $a-2$, then $a-4$ divides $a-2-(a-4)=2$ so $a-4\le2$ so $a\le6,$

so there is no solution with $a>6$.

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In other words, if $a-2=(a-4)c,$ then $2=(a-4)(c-1)$, so $a-4\le2,$ so $a\le6$.