Question

Find z, if it is known that:
\(a:-y^2+x^2=20\) 
\(b:y^3-2x^2-4z \geq -12\)and 
\(c: x,y\) and \(z\) are all positive integers

• A. Any integer greater than 0 and less than 24
• B. 24
• C. We need one more equation to find z
• D. 6
• E. 1

Answer 1

Answer: (E)

Step 1: From equation (a): 

\[ -y^2 + x^2 = 20 \quad \Rightarrow \quad x^2 = y^2 + 20 \]

Step 2: Substitute into equation (b):

\[ y^3 - 2x^2 - 4z \geq -12 \]

Since \(x^2 = y^2 + 20\):

\[ y^3 - 2(y^2 + 20) - 4z \geq -12 \]

\[ y^3 - 2y^2 - 40 - 4z \geq -12 \]

\[ y^3 - 2y^2 - 4z \geq 28 \]

Step 3: Trial with positive integers for \(y\):

• For \(y = 6\): \(x^2 = 36 + 20 = 56\) → not a perfect square.
• For \(y = 7\): \(x^2 = 49 + 20 = 69\) → not a perfect square.
• For \(y = 9\): \(x^2 = 81 + 20 = 101\) → not a perfect square.
• For \(y = 11\): \(x^2 = 121 + 20 = 141\) → not a perfect square.
• For \(y = 2\): \(x^2 = 4 + 20 = 24\) → not a perfect square.
• For \(y = 4\): \(x^2 = 16 + 20 = 36\) → \(x = 6\), valid!

Step 4: Substitute \(y = 4, x = 6\) into inequality:

\[ y^3 - 2x^2 - 4z \geq -12 \]

\[ 64 - 72 - 4z \geq -12 \]

\[ -8 - 4z \geq -12 \]

\[ -4z \geq -4 \quad \Rightarrow \quad z \leq 1 \]

Step 5: Since \(z\) is a positive integer:

\[ z = 1 \]

Final Answer: \(\boxed{1}\)