For every `a, b in N`
`a" @ "b=a^(2)" when "(a+b)" is even"`
`=a^(2)-b^(2)" when "(a+b)" is odd"`
`a#b=b^(2)" when "axxb" is odd"`
`=a^(2)-b^(2)" when "axxb" is even"`
What is the value of `(a#a)" @ "(a" @ "a)`?

Options

A. 0
B. `a^(4)`
C. either (a) or (b)
D. neither (a) nor (b)

Answer

To solve the problem, we need to evaluate the expression \((a \# a) \, @ \, (a \, @ \, a)\) using the defined operations for \( @ \) and \( \# \).

### Step 1: Calculate \(a \# a\)

According to the definition of the \( \# \) operation:
- If \( a \times a \) is odd, then \( a \# a = a^2 \).
- If \( a \times a \) is even, then \( a \# a = a^2 - a^2 = 0 \).

Since \( a \times a = a^2 \) and \( a^2 \) is always even for any natural number \( a \), we have:
\[ a \# a = 0 \]

### Step 2: Calculate \(a \, @ \, a\)

Next, we evaluate \( a \, @ \, a \):
- If \( a + a \) is even, then \( a \, @ \, a = a^2 \).
- If \( a + a \) is odd, then \( a \, @ \, a = a^2 - a^2 = 0 \).

Since \( a + a = 2a \) is always even for any natural number \( a \), we have:
\[ a \, @ \, a = a^2 \]

### Step 3: Combine the results

Now we substitute the results from Steps 1 and 2 into the expression \((a \# a) \, @ \, (a \, @ \, a)\):
\[ (a \# a) \, @ \, (a \, @ \, a) = 0 \, @ \, a^2 \]

### Step 4: Evaluate \(0 \, @ \, a^2\)

Now we evaluate \(0 \, @ \, a^2\):
- Since \( 0 + a^2 \) is even (as \( a^2 \) is even), we use the first case of the \( @ \) operation:
\[ 0 \, @ \, a^2 = 0^2 = 0 \]

### Final Result

Thus, the value of \((a \# a) \, @ \, (a \, @ \, a)\) is:
\[ \boxed{0} \]