ziyuenlau wrote:If a^2 is an integer and √(a^6−a^4−b−1)=10, what are the values of a^2 and b?
(1) a^4=a^2+20
(2) Product of b and b−1 is an even prime number.
Hi ziyuenlau,
We are given that √(a^6−a^4−b−1)=10 => a^6−a^4−b−1 = 100.
S1: a^4=a^2+20
a^4 - a^2 - 20 = 0
=> a^4 - 5a^2 + 4a^2 - 20 = 0
=> a^2(a^2 - 5) + 4(a^2 - 5) = 0
=> a^2 = 5 or -4.
Since a^2 = -4 implies that a is an imaginary number, thus we can discard it. So, a^2 = 5.
At a^2 = 5,
a^6−a^4−b−1 = 100 => 5^3 - 5^2 - b - 1 = 100 => 125 - 25 - b - 1 = 100 => b = -1.
So, a^2 = 5 and b = -1. Sufficient.
S2: Product of b and b−1 is an even prime number.
=> b(b - 1) = 2
=> b = -1 or 2.
At b = -1,
a^6−a^4−b−1 = 100 => a^6−a^4−(-1)−1 = 100 => a^6 - a^4 = 100.
Say X = a^2
=> X^3 - X^2 = 100
=> X^2(X - 1) = 100
=> X^2(X - 1) = (5^2)*4
=> X^2(X - 1) = (5^2)*(5 - 1)
=> X = a^2 = 5.
At b = 2,
a^6−a^4−b−1 = 100 => a^6−a^4−2−1 = 100 => a^6 - a^4 = 103.
It's difficult to calculate the value of a^2 since 103 is a prime number. Even if we calculate it, it would not be an integer. An optimum way to test is: plug-in a^ = 4 and a^2 = 6 [I proposed to test these values as at a^2 = 5, we get RHS = 100] and see if the RHS ( a^6 - a^4) of the equation a^6 - a^4 = 103 returns 103. You would find that it does not, so we have unique values of a^2 = 5 and b = -1. Sufficient.
The correct answer:
D
Hope this helps!
Relevant book:
Manhattan Review GMAT Data Sufficiency Guide
-Jay
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