## Wednesday, December 19, 2012

### Equations without Whole Number Solutions - Fruitless Search?

Hi everyone! Hopefully you are fine today. Not long until 2013. And yes, I am 99.44% confident that the human race will still be on Earth be here come 12/22/2012.

x + x^2 = n^2

Let n,x ∈ N. N represents the natural numbers. Natural numbers are commonly referred to the whole numbers 1, 2, 3, etc. Some mathematicians include 0.

Are there any integer solutions to x + x^2 = n^2 with n,x ∈ N?

Our first instinct is most likely to go grab the nearest calculator or computer. Observe that:

(I). x + x^2 = n^2

(II). x^2 * (1/x + 1) = n^2

Taking the (principal) square root of both sides yields:

(III). x * √(1/x + 1) = n

In order for (III) to be true:
1. The quantity 1/x + 1 has to be an integer,
2. 1/x + 1 has to be a perfect square, and
3. x = 1/x + 1

The only natural number that allows condition 1 to be true is when x = 1.

Then 1/x + 1 = 2.

We know that 2 is not a perfect square (√2 ≈ 1.41421), so condition 2 fails. (II) does not fit because 1 * √2 = √2, leaving n = √2, not fitting the requirement that n,x ∈ N.

According to this analysis, there are no natural number solutions to x + x^2 = n^2.

x + x^2 + x^3 = n^q

Let n,x ∈ N. Suppose q = 2. Can we find solutions with these conditions?

x + x^2 + x^3 = n^2

x^2 * (1/x + 1 + x) = n^2

x * √(1/x + 1 + x) = n

Once again the only way 1/x + 1 + x is an integer is that when x = 1. However when x=1, 1/x + 1 + x = 3, and √3 ≈ 1.73205. And 1 * √3 = √3, n = √3, which is not a natural number.

There are no solutions (in the natural number set) for x + x^2 + x^3 = n^2.

Then x + x^2 + x^3 = n^3

x^3 * (1 + 1/x + 1/(x^2)) = n^3

x * ∛(1 + 1/x + 1/(x^2)) = n

Again, the only possible candidate is when x=1, but that leaves n = ∛3. No solutions in the natural number set.

Not all is "lost"...

Consider trying to find solutions to:

x + x^2 + x^3 + x^4 = n^2 where n,x ∈ N.

Then x * √(1/x + 1 + x + x^2) = n.

If x = 1, then √(1/x + 1 + x + x^2) = √4 = 2, and n = 1 * 2 = 2. Success! I believe x=1 and n=2 is the only solution to this equation with these conditions imposed.

Happy Holidays everyone and see you next time!

Eddie

This blog is property of Edward Shore. 2012