I have given $7n+3$ for all integers $n$. And $7n+3$ should never be the cube of an integer $m$. I would prefer a solution using modular arithmetic but I am not sure how to show this.

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    $\begingroup$ The cubes $\pmod 7$ are $0^3=0,1^3=1,2^3=1,3^3=-1,4^3=1,5^3=-1,6^3=-1$ $\endgroup$
    – lulu
    Nov 6, 2016 at 16:19
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    $\begingroup$ Compare with this MSE-question. $\endgroup$ Nov 6, 2016 at 19:52

2 Answers 2


Hint $\,\ {\rm mod}\ 7\!:\,\ \left[ a^{\large 3}\equiv 3\right]^{\large 2}\! \Rightarrow\, a^{\large 6}\equiv 3^2 \equiv 2\ $ contra little Fermat.

Remark $\ $ Responding to a comment, the bracketed notation used above to simultaneously square both sides of the congruence is not in wide use, so keep this in mind if you employ it.

  • $\begingroup$ I have never encountered this notation for exponentiation of both sides of an equation (or congruence in this case). Is this your own notation or did you find it somewhere? $\endgroup$
    – Wojowu
    Nov 6, 2016 at 20:55
  • $\begingroup$ @Wojowu I don't recall seeing it anywhere else. $\endgroup$ Nov 6, 2016 at 21:03
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    $\begingroup$ In my experience, this is certainly not standard notation. I find using words to be clearer here: "if $a^3\equiv3\pmod 7$, then squaring both sides would yield $a^6\equiv2\pmod 7$, which contradicts Fermat's little theorem". The primary goal in communication (even communication of math) is clarity, not brevity. $\endgroup$ Nov 6, 2016 at 21:41
  • $\begingroup$ @Greg I choose my notation very carefully in order to highlight the essence of the matter. Whether or not it is "standard" matters little in this regard. I've been using that notation for many decades and I don't recall a single question about its meaning. $\endgroup$ Nov 6, 2016 at 21:45
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    $\begingroup$ I think that notation is quite suggestive. It does "highlight the essence" precisely because it's not standard usage. But I wouldn't want a student in an elementary course somewhere springing on his or her teacher because they found it here. $\endgroup$ Nov 6, 2016 at 22:01

Since $$k^3\equiv 0,1,6 \mod 7,$$ we have $$7 \not \mid k^3-3.$$

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    $\begingroup$ could you give a little bit more detailed explanation please? $\endgroup$
    – Arji
    Nov 6, 2016 at 16:30
  • $\begingroup$ you can check the first comment below the question. It is same argument. $\endgroup$ Nov 6, 2016 at 16:45
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    $\begingroup$ It might help to reference (or prove) the first statement. As currently presented, it seems like a fact that comes out of nowhere. $\endgroup$
    – R.M.
    Nov 6, 2016 at 20:56
  • $\begingroup$ We are just taking cubes and getting reminder for k=0,1,2,3,4,5,6. These are basics and comes from basics. $\endgroup$ Nov 6, 2016 at 20:59
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    $\begingroup$ While the argument is no doubt clear and obvious to you, you would help your Readers and improve the Answer by filling in the requested explanation. Comments are pretty ephemeral in comparison to Answers, so strive to make the Answer self-contained (but feel free to credit the Community member whose Comment you found relevant). $\endgroup$
    – hardmath
    Nov 7, 2016 at 0:02

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