Question

Which two-digit number is equal to its square of sum?

Should the equation look like this?
`10a+b=(a+b)^2`
If the answer is yes, how to solve it?

Answer 1

`81`

Explanation:

If the tens digit is `a` and the units digit `b`, then `a, b` must satisfy:

`10a+b = (a+b)^2 = a^2+2ab+b^2`

Subtracting `10a+b` from both ends, this becomes:

`0 = a^2+2(b-5)a+b(b-1)`

`color(white)(0) = a^2+2(b-5)+(b-5)^2+(b(b-1)-(b-5)^2)`

`color(white)(0) = (a+(b-5))^2+(b^2-b-b^2+10b-25))`

`color(white)(0) = (a+(b-5))^2-(25-9b)`

So:

`a+b-5 = +-sqrt(25-9b)`

In order for `25-9b` to be a perfect square, we require `b=1`.

Then:

`a+b-5 = +-sqrt(25-9) = +-sqrt(16) = +-4`

So:

`a = 5-b+-4 = 4+-4`

So the only non-zero value for `a` is `a=8`.

We find:

`81 = 9^2 = (8+1)^2" "` as required.

Alternatively we could just have looked at the first few square numbers and checked:

`16 = 4^2 != (1+6)^2`

`25 = 5^2 != (2+5)^2`

`36 = 6^2 != (3+6)^2`

`49 = 7^2 != (4+9)^2`

`64 = 8^2 != (6+4)^2`

`81 = 9^2 = (8+1)^2" "` Yes.