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Prove that the system of equations has no real solutions:

$$\begin{cases} y=\sqrt{x+\sqrt{1-x}} \\ x=\sqrt{y-\sqrt{1+y}}\end{cases}. $$

This is a former problem from a national math contest which I´ve solved already. However, since my solution was very similar to the solution manual, I´m interested in other solving approaches.

My idea, in short, was as follows:

We first notice that from equation (1), we get the condition $y < \sqrt{2}$.

From equation (2) we have that $y \geq \sqrt{1+y}$, but for $y<1$ this can't be true since then $\sqrt{1+y}>1$. So $y \geq 1$ and thus $$0 \leq y-\sqrt{1+y}\leq y-\sqrt{2},$$ which is a contradiction to our first condition $y < \sqrt{2}$.

the_fox
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alot
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1 Answers1

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Suppose that there exists a set of real solution.

First of all, $x$ has to satisfy the followings :

$$1-x\ge 0\ \ \ \text{and}\ \ \ x+\sqrt{1-x}\ge 0\ \ \ \text{and}\ \ \ x\ge 0.$$

This is equivalent to $0\le x\le 1$.

Similarly, $y$ has to satisfy the followings :

$$1+y\ge 0\ \ \ \text{and}\ \ \ y-\sqrt{1+y}\ge 0\ \ \ \text{and}\ \ \ y\ge 0.$$ This is equivalent to $y\ge\frac{1+\sqrt 5}{2}$.

Now let $f(x)=\sqrt{x+\sqrt{1-x}}\ \ \ (0\le x\le 1)$. Since we have $$f'(x)=\frac{2\sqrt{1-x}-1}{4\sqrt{1-x}\cdot\sqrt{x+\sqrt{1-x}}},$$ we have $f'(x)=0\iff x=\frac 34$, and so $1=f(0)=f(1)\le f(x)\le f(3/4)=\frac{\sqrt 5}{2}$.

However, this contradicts $y\ge\frac{1+\sqrt 5}{2}$.

mathlove
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