Twitter Question Answers

To find the value of x, we need to use a bit of algebraic manipulation.

Let’s assume that 3x + 1 is equal to a perfect square, say y^2. This means that:

3x + 1 = y^2

Now we can rearrange this equation to get:

3x = y^2 – 1

3x = (y + 1)(y – 1)

Since x is a positive integer, y must also be an odd positive integer (because y^2 is odd and 3 is odd). This means that y + 1 and y – 1 are consecutive even integers.

The only way for 3x to be the product of two consecutive even integers is if x is itself an even integer. So let’s assume that x = 2n, where n is some positive integer.

If x = 2n, then we can rewrite the second equation as:

4(2n) + 1 = 8n + 1 = z^3

Where z is some positive integer. Now we can solve for n by using trial and error or by observing that 8n+1 is one less than a multiple of 9, so it must be a perfect cube only if n=1. Therefore:

n = 1, x = 2n = 2, y^2 = 3x + 1 = 7, and z^3 = 4x + 1 = 9.

So the value of x is 2.