Find the sum of all positive integers b such that b^2 =a^3 +1 , where a is a prime number
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To find the sum of all positive integers b such that b^2 = a^3 + 1, where a is a prime number, we can use a systematic approach.
First, let's consider the equation b^2 = a^3 + 1.
We know that a prime number is only divisible by 1 and itself. So, a^3 is always divisible by a itself. Hence, a^3 + 1 will always result in a remainder of 1 when divided by a.
Now, let's examine the equation modulo a. Taking both sides modulo a, we get:
b^2 ≡ 1 (mod a)
This means that b^2 leaves a remainder of 1 when divided by a. So, b^2 - 1 is divisible by a.
Using difference of squares, we can rewrite b^2 - 1 as (b - 1)(b + 1). So, (b - 1)(b + 1) is divisible by a.
Since a is a prime number, it can divide either (b - 1) or (b + 1) but not both.
Considering the two possibilities:
1. If a divides (b - 1), we can write b - 1 = ak, where k is an integer. This implies b = ak + 1.
2. If a divides (b + 1), we can write b + 1 = ak, where k is an integer. This implies b = ak - 1.
Next, we need to substitute these values of b in the original equation b^2 = a^3 + 1 and solve for suitable values of k.
For the first case, substituting b = ak + 1 in the equation gives:
(ak + 1)^2 = a^3 + 1
a^2k^2 + 2ak + 1 = a^3 + 1
Simplifying, we get:
a^2k^2 + 2ak = a^3
a(a-1)k^2 + 2k = a^2
Since a is a prime number, it cannot divide both a and a-1. So, we have two sub-cases to consider:
Sub-case 1: If a divides a-1, we can write a-1 = ac, where c is an integer. This implies a = ac + 1.
Sub-case 2: If a divides a, we can write a = ac, where c is an integer.
For Sub-case 1, substituting a = ac + 1 in the equation a(a-1)k^2 + 2k = a^2, we get:
(ac + 1)(ac)k^2 + 2k = (ac + 1)^2
ac(ac+1)k^2 + 2k = (ac + 1)^2
Expanding, we get:
a^2c^2k^2 + ack^2 + 2k = a^2c^2 + 2ac + 1
Simplifying further, we get:
ack^2 + 2k = ac^2 + 2c + 1
This equation allows us to find suitable values of k for each prime a that satisfies the initial condition b^2 = a^3 + 1. We can iterate through different values of a and solve this equation to find the corresponding values of k.
For Sub-case 2, substituting a = ac in the equation a(a-1)k^2 + 2k = a^2, we get:
ac(ac-1)k^2 + 2k = a^2c^2
ac(ac-1)k^2 + 2k = a^2c^2
Expanding, we get:
a^2c^2k^2 - ack^2 + 2k = a^2c^2
Simplifying further, we get:
ack^2 - 2k = 0
This equation has a trivial solution k = 0. However, we are looking for positive integers b. So, let's continue with the previous cases.
Now, we can use the values of k obtained from the two sub-cases to find the corresponding values of b.
For each prime a, substitute the values of k obtained into the respective equations b = ak + 1 and b = ak - 1 to find the corresponding values of b.
Finally, sum up all the positive integers b obtained and that will be the solution to the initial problem.