If `(3+asqrt2)^100+(3+bsqrt2)^100=7+5sqrt2` number of pairs (a, b) for which the equation is true is, (a, b are rational numbers) (a) `1` (b) `6` (c) `0` (d) infinite

Options

A. `1`
B. `6`
C. `0`
D. infinite

Explanation

To solve the problem, we need to analyze the given equation:

\[ (3 + a\sqrt{2})^{100} + (3 + b\sqrt{2})^{100} = 7 + 5\sqrt{2} \]

### Step 1: Apply the Binomial Theorem
Using the Binomial Theorem, we can expand both terms on the left side.

\[ (3 + a\sqrt{2})^{100} = \sum_{k=0}^{100} \binom{100}{k} 3^{100-k} (a\sqrt{2})^k \]

This can be separated into even and odd powers of \(\sqrt{2}\):
- Even powers contribute to the rational part.
- Odd powers contribute to the irrational part.

### Step 2: Separate Even and Odd Powers
The expansion can be expressed as:

\[ (3 + a\sqrt{2})^{100} = \text{(even terms)} + \text{(odd terms)}\sqrt{2} \]

Similarly, for \((3 + b\sqrt{2})^{100}\):

\[ (3 + b\sqrt{2})^{100} = \text{(even terms)} + \text{(odd terms)}\sqrt{2} \]

### Step 3: Combine the Two Expansions
Combining both expansions gives us:

\[ \text{(even terms from both)} + \text{(odd terms from both)}\sqrt{2} = 7 + 5\sqrt{2} \]

### Step 4: Set Up Equations
From the above, we can set up two equations:

1. The sum of the even terms must equal 7.
2. The sum of the odd terms must equal 5.

### Step 5: Analyze the Even Terms
The even terms can be expressed as:

\[ \sum_{k \text{ even}} \binom{100}{k} 3^{100-k} (a^k + b^k) \]

### Step 6: Analyze the Odd Terms
The odd terms can be expressed as:

\[ \sum_{k \text{ odd}} \binom{100}{k} 3^{100-k} (a^k + b^k) \]

### Step 7: Solve for Rational \(a\) and \(b\)
To satisfy both equations, we need to find rational numbers \(a\) and \(b\) such that:

1. The sum of the even powers equals 7.
2. The sum of the odd powers equals 5.

### Step 8: Determine Feasibility
After analyzing the equations, we find that there are no rational pairs \((a, b)\) that can satisfy both conditions simultaneously.

### Conclusion
Thus, the number of pairs \((a, b)\) for which the equation holds true is:

\[ \boxed{0} \]