SOLUTION: Let (2 + \sqrt{5})(137) = a + b \sqrt{5}, where a and b are integers. Compute a^2

SOLUTION: Let (2 + \sqrt{5})(137) = a + b \sqrt{5}, where a and b are integers. Compute a^2 - 5b^2.

Question 1209707: Let (2 + \sqrt{5})(137) = a + b \sqrt{5}, where a and b are integers. Compute a^2 - 5b^2.

Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Let the given equation be
$$(2 + \sqrt{5})(137) = a + b\sqrt{5}$$
where $a$ and $b$ are integers.
Expanding the left side, we have
$$2(137) + 137\sqrt{5} = a + b\sqrt{5}$$
$$274 + 137\sqrt{5} = a + b\sqrt{5}$$
Comparing the rational and irrational parts, we get $a = 274$ and $b = 137$.
We are asked to compute $a^2 - 5b^2$.
$$a^2 - 5b^2 = (274)^2 - 5(137)^2 = (274)^2 - 5(137)^2$$
$$= (2 \cdot 137)^2 - 5(137)^2 = 4(137)^2 - 5(137)^2 = (4-5)(137)^2 = -1(137)^2 = -18769$$
Therefore, $a^2 - 5b^2 = -18769$.
Final Answer: The final answer is $\boxed{-18769}$

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