# 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.

Note by Zaid Baig
3 years, 9 months ago

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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{align} \\ (x+y)^2&=x^2+2xy+y^2 \\&\equiv y^2 \\&\equiv 1 \pmod{x} \\ \end{align} 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{align} \\ (x+y)^2&=(5+7)^2 \\&=12^2 \\&=144 \\&\equiv 4 \\&\not \equiv 1\pmod{5} \\ \end{align}

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{align} (x+y)^2&=(6+5)^2 \\&=11^2 \\&=121 \\&\equiv 1 \pmod{6} \end{align}

- 3 years, 9 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

- 3 years, 9 months ago