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

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.

• I think the problem means consecutive integers in base $10$. Anyway, $(x+1)^2-x^2=2x+1$ (you forgot $+1$). Try again after fixing this mistake and in case let me know if you need the solution Commented Jul 10 at 12:45
• Numbers $45$ and $55$, are in base $10$ or in base $b$? Clarify this point. Commented Jul 10 at 12:55
• I quoted the question as it is written in the book. $45$ and $55$ are in base $b$. My understanding is that the number requested (i.e $x$) is in base $b$ But from the answer of @ChrisLewis i see that $x$ it is most probably in base 10. Commented Jul 10 at 14:06

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.

• Simpler: subtract them $\Rightarrow b=2x\!+\!1,$ plugged into first $\Rightarrow 0=x^2\!-\!8x\!-\!9=(x\!-\!9)(x\!+\!1)\ \$ Commented Jul 10 at 16:51

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 $$b=19$$.