How to prove that $(X^y - X) \bmod 3 = 0$ always when $y$ is odd and $y > 1$, and $X >1$?

Asked Mar 05 '21 at 20:24

Active Mar 05 '21 at 20:43

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My question is quite simple: how to prove that $(X^y - X) \bmod 3 = 0$ when $y$ is odd and $y > 1$, and $x $ is an integer greater than $1$?

Examples:

$(2^3-2)/3=2$

$(11^3-11)/3=440$

$(7^5-7)/3=5600$

$(49^3-49)/3=39200$

$(13^7-13)/3=20916168$

Not always true for when $y=2$ (or other even numbers) examples:

$(5^2-5)/3 = 6.6666666\ldots$

$(5^4-5)/3 = 206.666666\ldots$

$(11^2-11)/3 = 36.666666\ldots$

edited Mar 05 '21 at 20:43

asked Mar 05 '21 at 20:24

Isaac Brenig

Modulo $6$ is redundant. All that's needed is modulo $3$ and this should be straightforward. – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Mar 05 '21 at 20:25
@TheSimpliFire. Just fixed it. Thanks – Isaac Brenig Mar 05 '21 at 20:39
Why do you have $X^y-X$ in the subject line but then $x^y-x$ in the body of the question? – Michael Hardy Mar 05 '21 at 20:41
Consider the cases $X=3x,3x+1,3x+2$. Modulo $3$ this reduces $X^y-X$ to $0^y+0,1^y+1,2^y+2$ respectively. The first two cases are proved immediately. For the last case note that $2^y+2\equiv(-1)^y-1\pmod3$. – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Mar 05 '21 at 20:42
Special case $,n = 3,,$ direction $(\Leftarrow)$ of the Theorem linked in the dupe (whose proof is very simple in that direction). – Bill Dubuque Mar 05 '21 at 20:49
Thanks @Bill, for the closure! – amWhy Mar 05 '21 at 21:12
Thanks @Bill Yes it answers my question – Isaac Brenig Mar 05 '21 at 21:18
Have you notice the remainder is either $0$ or $2$ it is never $1$? Note: we are never considering the remainder of $x^M$ without also considering the remainder of $x$. If the remainder of $x$ is $r$ where is the remainder of $x^M$ and what is the remainder of $x^M - x$. Also note: $x^M-x = x(x^{M-1} - 1)$ what does that tell you. – fleablood Mar 06 '21 at 00:46

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