Square of One more than the Larger Part This isn't really much of an algebra problem but different people are giving me different interpretations of " Square of One more than the Larger Part ". I've been given
" Separate 17 into two parts. So that the sum of their squares is equal to the "SQUARE OF ONE MORE THAN THE LARGER PART". Find the Smaller Part. "
I interpreted it myself; and I got

$x^2 + y^2 = (y^2 + 1)^2 = 17$

I don't think this works. I feel I have misinterpreted the problem. Other people have other interpretations too but they don't work either.
The correct answer is 5.

"Separate $17$ into two parts."
That tells me you should set up $x+y = 17$ (equation 1)
"The sum of the squares of the parts is equal to the square of one more than the greater part."
We can let $y$ be the greater part without any loss of generality (which makes $x$ the smaller part). That means that $x^2 + y^2 = (y+1)^2$. (equation 2).
Now you have to solve the two equations simultaneously.
Start by manipulating equation 2:
$x^2 = (y+1)^2 - y^2 = 2y+1$
$y = \frac{1}{2}(x^2 - 1)$
Substitute that expression for $y$ into equation 1 to get:
$x + \frac{1}{2}(x^2 - 1) = 17$
After rearrangement, you should get the quadratic:
$x^2 + 2x - 35 = 0$
which can be factorised:
$(x+7)(x-5) = 0$
So $x=-7$ or $5$.
Here we generally discard the negative root as splitting a number into two parts generally refers to natural number parts. Perhaps this needs better clarification in the question. But assuming this is the case, the only admissible solution is $x=5$.
