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Please check if my solution is correct or not :-

Since p is a prime number and greater than 3 therefore it will be of form $6k+1$ or $6k-1$

$Case 1:-$ When $p=6k+1$

$(6k+1)^3+(6k+3)=6k+4$

$18k+4/10/16$

$Case 2:-$ When $p=6k-1$

$(6k-1)^3+(6k+3)=6k+2$

$18k+2/8/14$

Therefore there are 6 different remainders 2,4,8,10,14,16 possible

Fin27
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    And have you found examples of each of those? – lulu Mar 24 '23 at 20:34
  • Examples are always key. What do you get if you take $p=5$? – lulu Mar 24 '23 at 20:37
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    Wouldn't $6k+2$ correspond to $18k+2/\color{red}{8/14}$? – J. W. Tanner Mar 24 '23 at 20:40
  • $$(6k \pm 1)^3 = 216k^3 \pm 108k^2 + 18k \pm 1 \equiv \pm 1 \pmod{18}.$$ – user2661923 Mar 24 '23 at 20:43
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    @user2661923 By $\rm \mu LTE$: $,p \equiv \pm1\pmod{2\cdot \color{#c00}3}\Rightarrow p^{\large \color{#c00}3}\equiv (\pm 1)^{\large \color{#c00}3}\pmod{2\cdot\color{#c00}3^2},\ $ so $,p^3+15\equiv 14,16\pmod{18}\ \ $ – Bill Dubuque Mar 24 '23 at 20:54
  • @BillDubuque Yes, I know that. However, the OP (i.e. original poster) used $~\pmod{6}~$ arithmetic. I am trying to keep things simple, following through on the specific approach that the OP attempted, while illustrating that the OP should be engaging in $~\pmod{18}~$ arithmetic. So, multiplying this out makes it easier for the OP to grasp the flaw in his approach. – user2661923 Mar 24 '23 at 20:59
  • Use direct calculation (using linked congruence laws) as in user's comment, or use $\rm\mu LTE$ as in my prior comment. For a solution-verification question to be on topic you must specify precisely which step in the proof you question, and why so. This site is not meant to be used as a proof checking machine. – Bill Dubuque Mar 24 '23 at 21:02
  • Note that the the proofs work for any integer $,p\equiv \pm1 \pmod{6},,$ i.e. integers coprime to $2$ & $3,,$ so it is not really about primes $> 3\ \ $ – Bill Dubuque Mar 24 '23 at 21:06
  • @user2661923 There's no need to compute all those coefficients when expanding the cube. rather simply note Binomial Theorem $\Rightarrow (i!+!6n)^3 = i^3 + \color{#c00}3\cdot i^2\cdot \color{#c00}6n + (\color{#c00}6n)^{\large\color{#c00}2}(\cdots)\equiv i^3\pmod{!\color{#c00}{18}}.,$ Put $,i = \pm1.\ \ $ – Bill Dubuque Mar 24 '23 at 21:21
  • @BillDubuque Again, I understand that. However, I am trying to keep things simple for the original poster. The idea is to illustrate the idea, as simply as possible, to someone brand new to number theory. The idea is not to try to force the OP into immediately stretching his intuition in a way that the OP might not be immediately comfortable doing. – user2661923 Mar 24 '23 at 21:24
  • @user2661923 But likely the point of the exercise is to exemplify these generalizations, and they will be obscured by using brute force computation (vs. intuition to simplify such). – Bill Dubuque Mar 24 '23 at 21:26
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    @BillDubuque That may have been the problem composer's point. My point was to communicate to the original poster. The communication should be shaped to the target audience. The ideas that you are broaching, are ideas that the OP may confront only after confronting the idea that he should be working in $\pmod{18}~$ rather than $~\pmod{6}. ~$ One step at a time. – user2661923 Mar 24 '23 at 21:29
  • @user2661923 I can't understand the fault in my solution , I understand what you did by expanding the whole equation, but what is the issue if I convert mod 6 to mod 18 ? Of all the 6 remainders 2,4,8,10,14,16 why only 14 and 16 survive ? – Fin27 Mar 24 '23 at 21:58
  • The fault in your approach is that you started out working in $~\pmod{6},~$ when the original problem involved $~\pmod{18}.~$ Attempting to convert an intermediate calculation from $~\pmod{6}~$ to $~\pmod{18}~$ is tricky, and should generally be avoided by a Math student new to Number Theory. ...see next comment – user2661923 Mar 24 '23 at 22:08
  • Your categorization of primes as $~(6k \pm 1)~$ is normally good. In this particular case, it is (still) good, because you can draw an immediate conclusion of the evaluation $~(6k \pm 1)^3, ~\color{red}{\pmod{18}}.~$ ...see next comment – user2661923 Mar 24 '23 at 22:10
  • Did you understand my first comment, where I multiplied out $~(6k \pm 1)^3~?~$ If so, do you agree with the Math? If so, then you know that you must have made a mistake in attempting the tricky conversion of $~\pmod{6}~$ to $~\pmod{18},~$ and you can use the Math in my comment as a guide to finding that mistake. Leave me another comment, if you continue to have questions. – user2661923 Mar 24 '23 at 22:13
  • Yes @user2661923 thanks a lot for the explanation , I do understand your first argument; The reason I used this intermediate calculation step was on account of a practice problem I saw prior to this particular question in which the author had used intermediate remainder to arrive at the answer ; Can I post that question and the solution the author has provided in the following comments, can you help me point out if there is an error in the author's solution as well ? – Fin27 Mar 24 '23 at 22:22
  • Wait. Let me set up a chat room, and send you a comment that contains a link to the chat room. When you get the comment, simply click on the link, and we can chat in private. – user2661923 Mar 24 '23 at 22:23
  • sure @user2661923, thank you ! – Fin27 Mar 24 '23 at 22:25
  • Here is the Chat room link. – user2661923 Mar 24 '23 at 22:27

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