Neat Proof for the statement: $\forall n\in\mathbb N.\ (n-1)(n+2) \not\equiv 2 \ \ \text{(mod 7)}$

71 Views Asked by At

I was trying to prove the statement: $$(n-1)(n+2) \not\equiv 2 \ \ \text{mod 7}$$ for all natural numbers and I was wondering if there was a really neat way to go about proving this (and other questions like this in general)?

One way I used (rather shoddily) is to just try $n\in[0,6]$ and show that none of them equal 2 under mod 7. This will obviously repeat periodically which is a sufficient proof but not a very nice one. This wouldn't be very feasible if say i wanted to prove a statement of the form $j \not\equiv i \ \text{mod k}$ where $k$ was very large (without a computer ofc).

Any pointers on how one can approach this and/or what theorems I could look at? I'm not overly familiar with number theory. Is there something obvious I'm missing?

1

There are 1 best solutions below

1
On BEST ANSWER

$$(n-1)(n+2)\equiv 2\pmod{7} \tag{1}$$ is equivalent to $$ 4n^2+4n+1 \equiv 3\pmod{7} \tag{2} $$ or to: $$ (2n+1)^2 \equiv 3\pmod{7} \tag{3}$$ that is impossible since $3$ is not a quadratic residue $\!\!\pmod{7}$.