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)