If $$x^2+y^2=2015$$ find the maximum value of $$x^2+2xy-y^2$$
I have actually solved this question but as you see this question is apparently of number theory and I have solved this using trigonometry. I want to know about other methods of solving this question. If they include just number theory concepts then that will be even better. I am sharing my attempt as an answer below. Any contribution is greatly appreciated.
Let $x=\sqrt{2015}\cos a$ and $y=\sqrt{2015}\sin a$
Then we can say that $$x^2-y^2+2xy=2015(\cos2a+\sin2a)$$ $$=2015\sqrt2\sin\left(\frac{\pi}{4}+2a\right)$$ Hence $$\textrm{max}(x^2-y^2+2xy)=2015\sqrt2$$
Like Maximum value of an expression by equation
let $x=my$ and $$x^2+2xy-y^2=c$$
As $xy\ne0,$
$$\dfrac c{2015}=\dfrac{1+2m-m^2}{1+m^2}$$
$$\iff m^2(c+2015)-4030m+c-2015=0$$
As $m$ is real, $$4030^2\ge4(c^2-2015^2)\iff c^2\le2(2015^2)$$
$$-2015\sqrt2\le c\le2015\sqrt2$$
The method you have used is the shorter one and easier too.
Alternative( using calculus) :
$f( x, y) =x^2+2xy-y^2$
and $g(x, y) =x^2+y^2-2015$
Let $\lambda$ be a parameter.Then by solving $\nabla f=\lambda\nabla g$ and $g(x, y) =0$ , we get points of extrema of $f(x,y)$ on $g(x,y)=0$.
Then we can check the function values of $f$ at points of extrema to determine maxima and minima.
This method is known as Lagrange Multipliers method.
Let's determine what condition shall $\lambda$ satisfy if $x^2+2xy-y^2\le \lambda(x^2+y^2)$.
$x^2+2xy-y^2\le \lambda(x^2+y^2)\Longleftrightarrow (\lambda-1)x^2-2xy+(\lambda+1)y^2\ge 0$. It is necessary that $\lambda\ge 1$, otherwise $(x,y)=(1,0)$ will make $LHS<0$. It is also necessary that $\forall y\ne 0$, $(\lambda-1)x^2-2xy+(\lambda+1)y^2\ge 0$, i.e. $ (\lambda-1)(\frac{x}{y})^2-2(\frac{x}{y})+(\lambda+1)\ge 0$
i.e. $\forall t\ne0$ $(\lambda-1)t^2-2t+(\lambda+1)\ge 0$. Let $\Delta>0$ we yield $\lambda^2\ge 2$. Combine this with $\lambda\ge 1$ we have $\lambda\ge\sqrt{2}$, which is also sufficient for it passes the $y=0$ test $\Longrightarrow x^2+2xy-y^2\le \sqrt{2}(x^2+y^2)=2015\sqrt{2}$. I shall leave it to you to check the equality can hold.
Solve for $y$:$$y=\pm\sqrt{2015-x^2}$$Substitute into the second expression: $$x^2+2xy-y^2=2x^2+2x\sqrt{2015-x^2}-2015$$Take the derivative of this function and set it equal to zero: $$4x+2\sqrt{2015-x^2}-\frac{2x^2}{\sqrt{2015-x^2}}=0\implies2\sqrt{2015-x^2}-\frac{2x^2}{\sqrt{2015-x^2}}=-4x$$Square both sides: $$4030-24x^2+\frac{4x^2}{2015-x^2}=0\implies4030*2015-4030x^2-24*2015x^2+24x^4+4x^2=0$$Make the substitution $v=x^2$ and you will get two solutions. Substitute them back into $x^2+2xy-y^2$ and the one that gives the bigger answer is the answer.
Geometrically, the extrema happen when when $x^2+2xy-y^2=c$ is tangent to $x^2+y^2=2015$. So, by using the "$-\frac{F_x}{F_y}$-rule", $-\frac{2x+2y}{2x-2y}=-\frac{2x}{2y}$ and hence, $y^2+2xy-x^=0$ from which we get $$y=(-1\pm\sqrt2)x.$$ If $y=(\sqrt2-1)x$, then from circle equation, $(4-2\sqrt2)x^2=2015$ and hence $x^2+2xy-y^2=2015\sqrt2$. If $y=(-1-\sqrt2)x$, then from circle equation, $(4+2\sqrt2)x^2=2015$ and hence $x^2+2xy-y^2=-2015\sqrt2$.
