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

Question

asked Jan 18, 2020 219k views

1 Answer

Ethan Harris · Answer 1 · 2020-01-22T05:48:18+0000

Let
and
x+1 be these consecutive integers. For some base
, 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)