Find the number of real ordered pair(s) (x, y) for which: `16^(x^2+y) + 16^(x+y^2) = 1`

Options

A. 0
B. 1
C. 2
D. 3

Answer

To find the number of real ordered pairs \((x, y)\) that satisfy the equation:

\[ 16^{(x^2 + y)} + 16^{(x + y^2)} = 1 \]

we will follow these steps:

### Step 1: Rewrite the equation using properties of exponents
We can express \(16\) as \(4^2\), so we rewrite the equation as:

\[ (4^2)^{(x^2 + y)} + (4^2)^{(x + y^2)} = 1 \]

This simplifies to:

\[ 4^{2(x^2 + y)} + 4^{2(x + y^2)} = 1 \]

### Step 2: Let \(a = 4^{(x^2 + y)}\) and \(b = 4^{(x + y^2)}\)
Now we can rewrite the equation as:

\[ a^2 + b^2 = 1 \]

### Step 3: Apply the Arithmetic Mean-Geometric Mean Inequality (AM-GM)
According to the AM-GM inequality, we know that:

\[ \frac{a^2 + b^2}{2} \geq \sqrt{a^2 b^2} \]

This implies:

\[ a^2 + b^2 \geq 2ab \]

### Step 4: Substitute \(a^2 + b^2 = 1\)
From our equation, we have:

\[ 1 \geq 2ab \]

This leads to:

\[ ab \leq \frac{1}{2} \]

### Step 5: Express \(ab\) in terms of \(x\) and \(y\)
We know:

\[ ab = 4^{(x^2 + y + x + y^2)} = 4^{(x^2 + y^2 + x + y)} \]

### Step 6: Set up the inequality
Thus, we have:

\[ 4^{(x^2 + y^2 + x + y)} \leq \frac{1}{2} \]

Taking logarithms base 4 on both sides:

\[ x^2 + y^2 + x + y \leq -\frac{1}{2} \]

### Step 7: Rearranging the inequality
Rearranging gives:

\[ x^2 + y^2 + x + y + \frac{1}{2} \leq 0 \]

### Step 8: Completing the square
We can complete the square for \(x\) and \(y\):

\[ \left(x + \frac{1}{2}\right)^2 - \frac{1}{4} + \left(y + \frac{1}{2}\right)^2 - \frac{1}{4} \leq 0 \]

This simplifies to:

\[ \left(x + \frac{1}{2}\right)^2 + \left(y + \frac{1}{2}\right)^2 \leq \frac{1}{2} \]

### Step 9: Analyzing the result
The expression \(\left(x + \frac{1}{2}\right)^2 + \left(y + \frac{1}{2}\right)^2 = \frac{1}{2}\) represents a circle centered at \((-1/2, -1/2)\) with radius \(\sqrt{1/2}\).

### Step 10: Finding the points of intersection
The only way for the sum of two squares to equal zero (the minimum value) is if both squares are zero. Thus:

\[ \left(x + \frac{1}{2}\right)^2 = 0 \quad \text{and} \quad \left(y + \frac{1}{2}\right)^2 = 0 \]

This gives us:

\[ x + \frac{1}{2} = 0 \quad \Rightarrow \quad x = -\frac{1}{2} \]
\[ y + \frac{1}{2} = 0 \quad \Rightarrow \quad y = -\frac{1}{2} \]

### Conclusion
Thus, the only real ordered pair \((x, y)\) that satisfies the original equation is:

\[ \boxed{\left(-\frac{1}{2}, -\frac{1}{2}\right)} \]