# 1, 2, 3, 4, … , +1?

Did you know… that for any x, x(x+1)(x+2)(x+3) + 1 = k^2, where k is an integer related to x? That’s right. You’ve probably heard this before, and in this post, I’ll prove it.

I was looking through math textbooks, when I reached a section on factorisation. I looked through the sample questions… and I saw this. The question was:

What is the remainder when sqrt(2009*2010*2011*2012 + 1) is divided by 5?

I saw this, and I had a flashback. I remembered a theorem that I had seen but I didn’t understand why it was true. But when I saw the theorem in the book, I understood why.

Let’s focus first on the x(x+1)(x+2)(x+3). Group x and (x+3) together, and (x+1) and (x+2) together. Multiply within the groups, which will give x2 + 3x and x2+ 3x + 2. If you remember the difference of squares, a2 – 1 = (a+1)(a-1). Now… do you see it?

Let a = x2 + 3x + 1. x(x+3) is equivalent to a-1, and (x+1)(x+2) is equivalent to a+1. Multiplying them together gives a2 – 1, and adding one gives a2, or a square number. Therefore, the value of k is x2 + 3x + 1.

As for the answer to the original question:

k = 20092 + 3*2009 + 1, and there’s a simple method of finding the remainder for numbers in this style. 2009 mod 5 is equivalent to 4, and 4^2 = 16, 3*4 = 12, 16 + 12 + 1 = 29, and 29 mod 5 is equivalent to 4.

Yeah… that’s that.

I was supposed to write a “Best of 2021” post a couple of weeks back, but I didn’t know what to write in it and I didn’t have the time to write it, since my days were so filled with study. I really wanted it to be finished, and I’m sorry I couldn’t. I promise I’ll write a Favourites post in February, so I’ll see you then, unless I write a post beforehand, which will be unlikely because school starts for me soon.

Alright.

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