Is it possible to find an $n \in \mathbb{Z}$ s.t. $6 \mid (n^2-5)$

$n$ can have 6 possible values based on division algorithm: $6q, 6q+1,..., 6q+5, \exists q \in \mathbb {Z}.$

Need check only the remainder, as will it fall in the class of remainders given as follows:

The remainder of any number mod 6 will fall in the 6 bins: {$0,1, ..., 5$}. Taking the remainder of squares of these will have the values as: $0, 1, 4, 3, 4, 1$ respectively, or the set {$0, 1, 3, 4$}.

So, need find if that the value of $n^2-5 \mod 6$ will fall in the set. for this just need to take following steps:

(i) check the remainder of $n\mod 6 \implies \{0,1,...,5\}$

(ii) check the remainder of $n^2\mod 6 \implies \{0,1,4,3\}$

(iii) check the remainder of $(n^2-5)\mod 6$. I take that the remainder of $(n^2)\mod 6$ on subtraction by $5$ will simply take the subtraction as follows: {$-5, -4, -1, -2$} $\equiv$ {$1,2,5,4$}.

So, the question reduces to : if any of the four equations are possible:

a) $n^2 = 6$

b) $n^2 = 7$

c) $n^2 = 10$

d) $n^2 = 9$

Only the case (d) is possible, so the value of n for which $n\equiv 3\pmod 6$ are eligible. These are: $n =9$, or in general $9k$, $\forall k \in \mathbb {Z}$.

• You already have $n^2-5\equiv 1,2,5,4\pmod 6$ which implies that $n^2-5\not\equiv 0\pmod 6$. – mathlove Dec 1 '17 at 10:21
• Where do you get $6,7,10,9$ from? You simply need to check whether zero is in the admissible set for $n^2-5$, which you have already calculated. If it were in the set you'd be looking for an example, but it isn't so no such example can exist. – Mark Bennet Dec 1 '17 at 10:28
• @MarkBennet I equated $n^2 - 5$ to the values obtained, i.e. $\{1,2,5,4\}$. Regarding the conditions, both $n^2-5$ and the set $\{1,2,5, 4\}$ are the same as obtained by mod by $6$. – jiten Dec 1 '17 at 10:32

Since $$n^2-5 \not \equiv 0 \pmod 6$$

no such $n$ exists.

Notice that $n =9$ is not a solution, $n^2-5 = 76$ is not divisible by $6$ since it is not divisible by $3$.

• Please tell flaw in my method of forming quadratic equation. – jiten Dec 1 '17 at 10:25
• $\{1,2,5,4\}$ are all the values that the remainder can take. If $n^2 \equiv 9 \pmod 6$, $n^2-5 \equiv 4 \pmod 6$, $n^2-5$ is still not divisible by $6$. – Siong Thye Goh Dec 1 '17 at 10:42
• Please have a look at my post at : math.stackexchange.com/q/2675761/424260 – jiten Mar 4 '18 at 5:19

If $6 | n^{2} - 5$ then $n^{2} - 5 \equiv 0 \mod 6$, i.e. $n^{2} \equiv 5 \mod 6$. Take a look at the squares of the elements of $\mathbf{Z}/6\mathbf{Z}$ and you will see no such number exists.

It is easier to see that $(3n\pm 1)^2=9n^2\pm 6n+1\equiv 1 \bmod 3$. So if a square is not divisible by $3$ it leaves remainder $1$ on division by $3$.

Now $6|n^2-5$ means that $n^2=6m+5=3(2m+1)+2\equiv 2 \bmod 3$.

• But, why my approach failed is not clear to me at all. I agree that all are correct, but then my forming quadratic equations is wrong. – jiten Dec 1 '17 at 10:23
• I equated $n^2 - 5$ to the values obtained, i.e. $\{1,2,5,4\}$. Regarding the conditions, both $n^2-5$ and the set $\{1,2,5, 4\}$ are the same as obtained by mod by $6$. – jiten Dec 1 '17 at 10:35