A modulus rule that I found out, can anyone help me prove or disprove this?

For any number $x$ where $x>=2$ and any number $y$ which $y$ mod $x \ne 0$

$(x+y)^2$ mod $x$ is equal to 1

I'm not quite sure how this can be useful, or if people have discovered it, but who knows. Please tell me if people have, and I'm coding a program that'll test it up to about a 100 in x and 1000 in y. I'll be back with the results and the github link to the program.

#Remainder #Math #Modulus #Theorems #Whoknowswhatelse

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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Comments

The statement is false... let me disprove it...

Assume the statement is true, that is for any number $x\geqslant 2$ and for any number $y$ and $x\nmid y$ which is $y\not\equiv 0 \pmod{x}$ . $\begin{aligned} \\ (x+y)^2&=x^2+2xy+y^2 \\&\equiv y^2 \\&\equiv 1 \pmod{x} \\ \end{aligned}$ Hence $y^2-1$ must be divisable by $x$ , now we can conclude for any number $y$ and $x\nmid y$ , $x\mid y^2-1$ .

However, since $y^2-1=(y+1)(y-1)$ , by Euclid's lemma, we get either $x\mid y+1$ is true or $x\mid y-1$ is true, so the equivalence $(x+y)^2 \equiv 1\pmod{x}$ cannot hold if $x\mid y+1$ and $x\mid y-1$ are both false.

For example, assume $x=5$ and $y=7$ , we have satisfied the conditions $y\not \equiv 0 \pmod{x}$ and $x\geqslant2$ , here, $x\mid y+1$ and $x\mid y-1$ are both false. $\begin{aligned} \\ (x+y)^2&=(5+7)^2 \\&=12^2 \\&=144 \\&\equiv 4 \\&\not \equiv 1\pmod{5} \\ \end{aligned}$

Other pairs of counterexamples $\{x,y\}$ are $\{10,8\}$ , $\{6,13\}$ , $\{25,9\}$ ... (there are infinity of them!) This statement can be made true if you add one more condition: one of the numbers $y+1$ or $y-1$ must be divisable by $x$ .

Therefore, the corrected statement are as follows:

For any number $x\geqslant2$ , $y$ , $x\nmid y$ , if either $x\mid y+1$ or $x\mid y-1$ is true, then $(x+y)^2\equiv 1 \pmod{x}$

For example, assume $x=6$ , $y=5$ , we have satisfied $x\geqslant2$ , $x\nmid y$ , $x\mid y+1$ $\begin{aligned} (x+y)^2&=(6+5)^2 \\&=11^2 \\&=121 \\&\equiv 1 \pmod{6} \end{aligned}$

Kenneth Tan - 6 years, 10 months ago

Thanks, I came up with it in the car, so who knows. And I coded it, most of them didn't work. It's basically y if (y+1) or (y-1) mod x equals 1? Thanks

Zaid Baig - 6 years, 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$