The binomial expansion of $$(a+bx)^{n}$$ begins $$16-8x\dots$$ If $$a,b,n$$ are all integers, find the possible sets of values for $$a,b$$ and $$n$$.

If $$a, b$$ are integers, but $$n$$ is not, show that $$a=2^{k}$$ where $$k≥3$$ and find expressions for $$b$$ and $$n$$ in term of $$k$$.

$$(a+bx)^{n}=a^{n}(1+\frac{bx}a)^{n}=a^{n}(1+(\frac{nbx}a+…)$$

So $$a^{n}=16$$ and $$\frac{a^{n}nbx}a=-8x$$ which gives $$\frac{16nbx}a=-8$$ so $$\frac{nb}a=-\frac 1 2$$ and $$b=-\frac a {2n}$$

If $$a$$ and $$n$$ are both integers, either $$a=16$$ or $$a=4$$ or $$a=-4$$ or $$a=2$$

If $$a=16, n=1, b=-\frac{16}2=-8$$ which gives $$(16-8x)^{1}=16-8x$$

If $$a=\pm4, n=2, b=\mp\frac4{2\times2}=\mp1$$ which gives $$(4-x)^{2}$$ or $$(-4+x)^{2}$$

If $$a=2, n=4, b=-\frac2{2\times8}$$ which is not an integer so it does not work.

If $$n$$ is negative than $$a^{n}=16$$ gives a value of $$a$$ between $$0$$ and $$1$$, so not possible.

If $$a$$ is an integer and $$a^{n}=16$$ then $$=(2^{4})^{1/n}=2\frac4 n$$ is an integer. So $$n=\frac 4 k$$

This gives $$b=-\frac a{2n}=-\frac{2^{k}}{2\times\frac4 k}=-2^{k-3}k$$ which is an integer for $$k≥3$$

(We have already seen that $$a=4, 16(k=2,4)$$ give integer values of $$n$$ and $$b$$)

Alternative approach – use a spreadsheet to investigate using a list of integers for $$a$$ with columns for $$n=\frac{\log 16}{\log a}$$ and $$b=-\frac a{2n}$$. It becomes clear that the powers of $$2$$ give integers for $$b$$ and rational values for $$n$$. Then create a list of the powers of $$2$$ and their corresponding values of $$n$$ and $$b$$ for further insight into the problem.