Help with modular equation for division of $n^2$ with $n+1$

Question

I am having trouble understanding the solution for the following problem:

Show that when dividing n x n by n+1, the remainder is always 1.

I have attempted to solve it using modular equations:

$$n \times n \equiv r \pmod{n+1}$$

Which simplifies to:

$$n^2 \equiv r \pmod{n+1}$$

However, when trying different values for n, I am not getting the expected result of the remainder always being 1. Can anyone please explain or provide a step-by-step breakdown of how this solution works for different values of $n$? I would greatly appreciate any help or guidance on this matter.

Answer 1

One way to show that $n^2\equiv 1 \pmod{n+1}$ is to write $n^2$ as $$n^2=(n+1)(n-1) +{\color{red}1}$$

Here, you can see that the remainder of the division of $n^2$ by $n+1$ is $\color{red}1$.

Answer 2

$$n^2=[(n+1)-1]^2=(n+1)^2-2(n+1)+1\equiv 0^2-2\times 0+1\equiv 1 \quad \pmod {n+1}$$

Answer 3

Modular arithmetic seems to be slightly overkill. One can use long division as a very elementary approach. What I'm about to show is slightly related to the approaches shown in the other answers.

$$\frac{n^2}{n+1} = \frac{n^2 \color{red}{-1+1}}{n+1}$$

Now we split the fractions and use difference of squares to get,

$$\frac{(n-1)(n+1)}{n+1}+\frac{1}{n+1}$$

I hope it's clear from this point what to do next.