Solving for Valid Positive Integers in the Given Equation
In this article, we explore the method to find the valid positive integer solutions for axk that satisfy the given equation a4x24x-1-99k2. We present a step-by-step approach, including the factorization and manipulation of the equation, followed by a thorough analysis of the potential solutions.
Equation Rearrangement
First, let's rewrite the original equation to make it more factorable:
a4x24x-1-99 k2 can be rearranged as:
a2x22x-1/22-k299
To facilitate factorization, we introduce a new variable u, such that:
u a2x22x-1/2
Thus, the equation becomes:
u2-k299
This can be further factored as a difference of squares:
(u-k)(u k)99
Integer Solutions Analysis
We now need to ensure that both u-k and u k are integers. Let us assign:
u-kc
u kd
And, as a result, we get:
cd99
c, d in mathbb{Z}
From the above, we set c99/d where d in {1, 3, 9, 11, 33, 99}. This gives us the following pairs:
d1, c99 d3, c33 d9, c11 d11, c9 d33, c3 d99, c1Next, we solve for k:
u-kc Rightarrow u kc
u kd Rightarrow u d-k
Setting these equal gives:
u1 -k c u1, u1 k d u1
Rearranging, we get:
2k1d d2-99
Since d geq 11, we calculate:
d11, 2k11 112-99 Rightarrow 22k 22 Rightarrow k 1 d33, 2k33 332-99 Rightarrow 66k 891 Rightarrow k 15 d99, 2k99 992-99 Rightarrow 198k 9702 Rightarrow k 49Checking the Exponent
Now, we need to check the exponent of a for these values of k:
u d-k 11-1 10 Rightarrow a2x22x-1/2 10 Rightarrow 2x22x-1/2 10 Rightarrow 4x24x -1 2 Rightarrow 4x24x -3 0 u d-k 33-15 18 Rightarrow a2x22x-1/2 18 Rightarrow 2x22x-1/2 18 Rightarrow 4x24x -1 36 Rightarrow 4x24x -37 0 u d-k 99-49 50 Rightarrow a2x22x-1/2 50 Rightarrow 2x22x-1/2 50 Rightarrow 4x24x -1 100 Rightarrow 4x24x -101 0We seek a solution where the exponent of a is 1. However, none of the above equations result in the exponent being 1. Therefore, there are no valid positive integer solutions for axk that satisfy the given equation.
Conclusion: There is no integer solution for axk that satisfies the equation a4x24x-1-99k2.