A Problem (finally!)

by v_Enhance, Jun 26, 2010, 11:06 PM

I decided I need to actually put some problems on this blog to make it less trivial, so I'll start off with a semi-easy one :P

Problem 1. For two $x,y \in \mathbb C$, we have $x+y = 2N$ and $xy = 2N^2$. Compute $x^2+y^2$, and prove that $x,y \in \mathbb R$ if and only if $N=0$.
This post has been edited 2 times. Last edited by v_Enhance, Jun 27, 2010, 3:02 AM

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First part: $x^2+y^2=(x+y)^2-2xy=(2N)^2-2 \cdot 2N^2=4N^2-4N^2=\boxed{0}$.

Second part: First, we assume $N=0$ and try to prove that $x,y \in \mathbb{R}$. We know $xy=2N^2=0$, so either $x=0$ or $y=0$. If $x=0$, then $0+y=0$, so $y=0$. Similarly, if $y=0$, $x=0$. Thus, the only ordered pair that works is $(x,y)=(0,0)$, which is real, so we have proven the "if" part.

Next, we assume $x,y \in \mathbb{R}$. From above, $x^2+y^2=0$ and, by the Trivial Inequality, $x^2+y^2\ge 0$. Since we have equality, we know $x=y=0$. Thus, $2N=x+y=0$, so $N=0$. This completes the "only if" part and the proof. $\blacksquare$
This post has been edited 1 time. Last edited by PowerOfPi, Jun 27, 2010, 4:31 AM

by PowerOfPi, Jun 27, 2010, 4:19 AM

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