Question

2. Suppose (220)_b and (251)_b are the squares of consecutive integers. Determine b.

Answer 1

Let $x$ and $x+1$ be these consecutive integers. For some base $b$, we can write

$(x_{10})^2=220_b=2b^2+2b$

$((x+1)_{10})^2=251_b=2b^2+5b+1$

If

$x^2=2b^2+2b$

then

$x^2+1=2b^2+2b+1$

Now,

$x^2+2x+1=2b^2+5b+1$

$2x+2b^2+2b+1=2b^2+5b+1$

$\implies2x=3b$

Squaring both sides gives

$4x^2=9b^2$

and so

$4(2b^2+2b)=9b^2$

$\implies8b^2+8b=9b^2$

$\implies b^2-8b=0\implies\boxed{b=8}$

(because the base has to be non-zero)