$\frac{'ab'}{'ba'}=\frac{'bc'}{'cb'}$ Question: 

We write $'pq'$ to denote a two digit integer with tens digit $p$ and units digit $q$. For which values of $a,b$ aand $c$ are the two fractions $\frac{'ab'}{'ba'}$and$\frac{'bc'}{'cb'}$  equal and different from $1$?

My attempt:

So, write this using "normal algebra";$$\frac{10a+b}{10b+a}=\frac{10b+c}{10c+b}$$Multiply by the two denominators:$$(10a+b)(10c+b)=(10b+c)(10b+a)$$Multiply out:$$100ac+10ab+10bc+b^2=100b^2+10ab+10bc+ac$$ And we can subtract $(10ab + 10bc) $ from each side. We are left with:$$100ac+b^2=100b^2+ac$$ Which, by rearranging, we get $$99ac=99b^2$$which is equivalent to (if we divide by $99$) $$ac=b^2$$
  $$[a,b,c]\neq 0$$ Since they are all parts of the tens digit, and if they are $0$, then they will not be a two digit number, which is vital. $a \neq b$, since if they were $=$, the fraction would $= 1$, and neither can this relationship occur between $b \neq c$

This was all written down on paper by me. Can anybody help me finish this off? 

Looks good so far. Now you just have to write down all the ways that $b^2 = ac$ for $1 \leq a, b, c \leq 9$, $a \neq b$, $b \neq c$ : $$\begin{align*} 2^2 &= 1 \times 4 \\ 3^2 &= 1 \times 9  \\ 4^2 &= 2 \times 8 \\ 6^2 &= 4 \times 9 \end{align*} $$
Each of these gives two solutions; e.g. $2^2 = 1 \times 4$ gives $(a, b, c) = (1, 2, 4)$ or $(a, b, c) = (4, 2, 1)$. The criterion $b^2 = ac$ states that $a, b, c$ stand in a geometric progression; if the common ratio of this progression is $r$, then $'bc' = r \times {'ab'}$ and $'cb' = r \times {'ba'}$, so this should make intuitive sense.
