Find a base b in which $\left( 45 \right)_{b}$ and $\left( 55 \right)_{b}$ are squares of consecutive integers

Question

I started with

$$(i) \hspace{5 mm}\left( 55 \right)_{b}-\left(45 \right)_{b}=\left(10 \right)_{b}=\left( b \right)_{10}$$

$$(ii) \hspace{5 mm} \left(x+1 \right)_{b}^2- \left( x \right)_{b}^2=\left( 2x\right)_{b} $$

$$ \Rightarrow \left( 2x\right)_{b}=\left(10 \right)_{b} $$

$$ \Rightarrow \left( x\right)_{b}=\left(5 \right)_{b} $$

This means that $ b \gt 5$.

With trial and error i checked for :

$(i) \hspace{2mm} b=6 $ and $ 5^2-4^2 $

$(ii) \hspace{2mm} b=7 $ and $ 6^2-5^2 $

$(iii) \hspace{2mm} b=8 $ and $ 7^2-6^2 $

but none of the above was right.

Answer 1

We're looking to solve $$4b+5=x^2,\quad 5b+5=(x+1)^2$$ for integers $b$ and $x$.

Eliminating $x$ will lead to unpleasant algebra, so let's try eliminating $b$:

\begin{align} 20b+25&=5x^2\\ 20b+20&=4(x+1)^2\\ 5&=x^2-8x-4 \\ x^2-8x-9&=0 \\ (x-9)(x+1)&=0 \end{align}

With $x=-1$, we get $b=-1$, which doesn't work as a base. With $x=9$, we have $b=19$, which is the only valid solution.

Answer 2

A slightly different line of attack:

Let $4b+5=n^2$ so that $5b+5=(n+1)^2=n^2+2n+1$. Subtracting the first equation from the second gives

$b=2n+1\implies n=(b-1)/2$

So

$4b+5=(b-1)^2/4$

$16b+20=b^2-2b+1$

$b^2-18b-19=0$

which has one positive root $b=19$.