Long story short: While doing some simple math exercises, I came across one that seemed impossible. Days later I decided to search the web for it, and found out there was a typo, putting the $^2$ outside the parenthesis.
Anyway, the REAL exercise was quite easy, but it got me wondering if it was possible to resolve the one with the mistake. Is it even possible to prove it's impossible?
Well, here it goes:
Knowing that $(x + y)^2 - (x - y^2) = 384$, find $x \cdot y$.
On
I assume you want $x,y$ integers. You are asking to solve $x^2+2xy+2y^2-x=384$ We can use the quadratic equation to get $x=\frac 12\left(1-2y\pm \sqrt{(1-2y)^2-4(2y^2-384)}\right)=\frac 12\left(1-2y\pm \sqrt{1537-4y-4y^2}\right)=\frac 12\left(1-2y\pm \sqrt{1538-(1+2y)^2}\right)$ We must have $(1+2y )^2 \lt 1538$ or $-20 \le y \le 19$ for the square root to be real. We can then try all the $y$'s and find solutions for $(x,y)=(26,-19),(13,-19),(26,-7),(-11,-7),(13,6),(-24,6),(-11,18),(-24,18)$
$(13+6)^2-(13-6^2)=384$ gives an integer solution to $(x+y)^2-(x-y^2)=384.$ This would then give $xy=13\cdot 6=78.$ However there may be other integer solutions, and the problem doesn't make it clear whether only positive integers are OK, or even reals.
Added: If $a=x+y$ then we require also $x-y^2=384-a^2$ and then by subtraction that $y+y^2=384+a-a^2.$ This is $y(y+1)+a(a-1)=384,$ or $$(2y+1)^2+(2a-1)^2=1538 \tag{*}$$ after multiplying by $4$ ands completing the square. This makes it clear that if one is willing to use negative integers, there are at least three more solutions obtained by changing signs on the squares in $(*).$ One such solution is $x=-11,y=-7,$ and for this one, $xy=77$ doesn't match the $78$ from the other positive integer solution above (so using any integers the problem doesn't have a uniuque solution).
I worked out the other related solutions obtained from $(*)$ on changing signs on the squares. So in all I have $(x,y)=(13.6),\ (-11,-7),\ (26,-7),\ (-24,6)$ as integer solutions. That there aren't more can be seen from the facts that $1538$ is $2$ mod $8$ so only two odd squares could have that as a sum, and also $1538=2 \cdot 769$ with $769$ a prime which is $1$ mod $4$ and so has only one representation as a sum of squares $769=12^2+25^2.$ Working this through with the only factorization of $2$ as $(1+i)(1-i)$ (up to0 units), one doesn't get more than one representation of $1538$ as a sum of two squares.
Final note: As @jp26 has said in a comment there are more solutions obtained by switching the squares(which is which) on the right of $(*)$.