# Why every square number is of the form $$5n$$ or $$5n \pm 1$$

29 July 20

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I have recently started going through my maths notes again. This is one of my notes that were taken from my reading of Higher Algebra. The purpose of this post is to see if mathjax works with my script to generate the website, and whether the LaTeX syntax works with markdown, so it is purely a technical exercise, and a reference for future me on how I implemented it.

Let us say you are trying to divide one number by another. Let us use the example of dividing $$a$$ by $$b$$. We all know that doing this results in a quotient and a remainder of the form: $a = bq + r \quad \text{and} \quad 0 \leqslant r < b.$ Where $$q$$ is the quotient and $$r$$ is the remainder. From here, we can formulate another equation that shows that there is an $$r'$$ where: $r = b - r'$ And therefore we have: $a = b(q+1) - r' \quad \text{and} \quad 0 \leqslant r' < b;$ From this, we can see that if $$r \geqslant \frac{1}{2}b$$ then $$r' \leqslant \frac{1}{2}b$$. Therefore we know that there exists numbers $$Q$$ and $$R$$ such that: $a = bQ + R \quad \text{and} \quad \left|R\right| \leqslant \frac{1}{2}b$ We now have an equation to describe how far $$\left|R\right|$$ any number $$a$$ is from a multiple of any other number $$b$$. We can therefore focus now on the square numbers which will be of the form: $a^2 = \left(bQ +R\right)^2 \quad \text{and} \quad \left|R\right| \leqslant \frac{1}{2}b$ And since we are trying to find out how far all square numbers will be from multiples of five we can substitute $$b = 5$$: $a^2 = \left(5Q +R\right)^2 \quad \text{and} \quad \left|R\right| \leqslant 2$ Do note that instead of $$\frac{1}{2}b$$ being 2.5 it will be two since this equation is specifically defined to be discrete. Expanding the equation, we find: $a^2 = 5\left(5Q^2 + 2QR\right) + R^2 \quad \text{and} \quad \left|R\right| \leqslant 2$ We know that: $R = \{-2,-1,0,1,2\}$ And therefore: $R^2 = \{0,1,4\}$ So when we go back to the equation: $a^2 = \underbrace{5\left(5Q^2 + 2QR\right)}_\text{Divisible by 5} + \{0,1,4\}$ Now, let $$n = \left(5Q^2 + 2QR\right)$$, and we can show that: $a^2 = 5n + \{0,1,4\}$ Which are of the following forms:

\begin{align} a^2 & = 5n \\ a^2 & = 5n + 1 \\ a^2 & = 5(n+1) - 1 \\ \end{align}

Therefore proving that all square numbers are of the form $$5n$$ or $$5n \pm 1$$.

Best Regards,
Philip

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