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I've started to learn math not that long ago, what I actually must have been doing at school, but never mind.

Solving exercises, I ask myself pretty often, if I'm right making my final conclusions, proving a theorem. Of course, I haven't faced yet with something really hard to solve. Currently I'm not gonna ask about proofs for any type of exercises, but rather about my specific case I've faced with.

Just today I solved the next exercise:

Prove that for any natural value of "n"(not equal to 1), the result of the expression $(n^4 + n^2 + 1)$ is a composite number.

Not without help, but I solved it, getting the next result $$(n^2 + n + 1)(n^2 - n + 1)$$


And here is the point. I've proved it just by substituting several random numbers in the place of "n" variable, getting the right result. And this seems to be such a lame and a wrong approach, because there are ∞ numbers, when I've tried just 3-5 of them. And that's it.
How can I make sure that my proofs are right, not by substituting numbers and not by making something unreliable like substituting, but by... I don't know even by why. I'm confused about that.
I hope you can advise me something and give me some tips, maybe explanations, because possibly I'm wrong or just don't understand something important. What do you think ?

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  • $\begingroup$ Not all results that 'seems' correct proved a theorem. $\endgroup$
    – Eddy Piedad
    Commented Jun 22, 2022 at 12:31
  • $\begingroup$ Did you make the connection between the statements “$n^4+n^2+1$ is equal to $(n^2+n+1)(n^2-n+1)$ for all $n$” and “$n^4+n^2+1$ is a composite number for all integers $n>1$”? $\endgroup$
    – user700480
    Commented Jun 22, 2022 at 12:31
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    $\begingroup$ Not sure what you are asking. In this case you can multiply your two quadratics together and confirm that you get the given quartic. You are correct that working examples, on its own, may not lead to a proof but it's fine if you get the idea this way so long as you then complete the proof. $\endgroup$
    – lulu
    Commented Jun 22, 2022 at 12:32
  • $\begingroup$ Do you know how to multiply $n^2+n+1$ by $n^2-n+1$? The pedestrian method: multiply each-by-each, you get $9$ terms, some will cancel: $(n^2+n+1)(n^2-n+1)=n^2n^2-n^2n+n^2+n\cdot n^2-n\cdot n+n+n^2-n+1=n^4-n^3+n^2+n^3-n^2+n+n^2-n+1=n^4+n^2+1$. Advanced method: notice $n^2+n+1=(n^2+1)+n$ and $n^2-n+1=(n^2+1)-n$ so you are calculating $((n^2+1)+n)((n^2+1)-n)=(n^2+1)^2-n^2=n^4+2n^2+1-n^2=n^4+n^2+1$ (using identities $(a+b)^2=a^2+2ab+b^2$ and $(a+b)(a-b)=a^2-b^2$). $\endgroup$
    – user700480
    Commented Jun 22, 2022 at 13:11
  • $\begingroup$ ("Boss" method: $n^4+n^2+1$ and $(n^2+n+1)(n^2-n+1)$ are both polynomials in $n$ of degree $4$, so there is a theorem that states that, if they agree on $5=4+1$ points, they are the same polynomial. Thus, you can actually just substitute five different values of $n$, if they match, then they will match for any other value of $n$! This means, with enough heavyweight machinery your trial-and-error approach can be justified. You are, however, at your stage not expected to know about this theorem or know how to correctly use it.) $\endgroup$
    – user700480
    Commented Jun 22, 2022 at 13:18

1 Answer 1

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Proof making is a constructive process: in every new proof you use properties you proved before and combine them in a very clear way.

In this case you need two preceding facts

  1. $(a+b)(a-b) = a^2 - b^2$
  2. $(a + b)^2 = a^2 + 2ab + b^2$

which in turn are derived from more basic ones, as I will indicate below.

To illustrate the process I will go through it in a rather lengthy development. Note however that you should proceed incrementally, starting from very simple equations and progressing toward more complex ones, step by step, in separate exercises.

For the first property: \begin{align*} (a+b)(a-b) &= a(a-b) + b(a-b) &&\textrm{; distrib.}\\ &= (a^2 + a(-b)) + (ba + b(-b)) &&\textrm{; distrib.}\\ &= a^2 + (a(-b) + ba) + b(-b)) &&\textrm{; assoc. (twice)}\\ &= a^2 + (a(-b) + ab) + b(-b)) &&\textrm{; commut.}\\ &= a^2 + (a(-b + b)) + b(-b)) &&\textrm{; distrib.}\\ &= a^2 + (a0 + b(-b)) &&\textrm{; add. inv.}\\ &= a^2 + (0 + b(-b)) &&\textrm{; zero mult.}\\ &= a^2 + b(-b) &&\textrm{; zero sum}\\ &= a^2 - b^2 &&\textrm{; add. inv. mult.} \end{align*} Note that you should know why each of the justifications on the right end holds. If not, step back and prove them.

The second equation follows similarly so I will leave it to you as an exercise.

Now the property you are interested in: \begin{align*} (n^2 + n + 1)(n^2 - n + 1) &= ((n^2 + 1) + n)((n^2 + 1) - n) &&\textrm{; commut. + assoc.}\\ &= (n^2+1)^2 - n^2 &&\textrm{; fact 1 above}\\ &= (n^2)^2 + 2n^2+1 - n^2 &&\textrm{; fact 2 above}\\ &= n^4 + 2n^2 - n^2 + 1 &&\textrm{; commut.}\\ &= n^4 + n^2 + 1. \end{align*} So far we have proven $$ n^4 + n^2 + 1 = (n^2+n+1)(n^2-n+1). $$ However, this does not automatically imply that this is a composite number. We first need to show that $n^2+n+1>1$ and $n^2-n+1>1$, provided that $n>1$. So, we have to deduce these inequalities: \begin{align*} n^2+n+1 &> n^2 - n + 1 &&\textrm{; $n > -n$}\\ &= n(n-1) + 1 &&\textrm{; distrib.}\\ &\ge n1 + 1 &&\textrm{; $n>1\Rightarrow n-1>0\Rightarrow n-1\ge1$}\\ &= n + 1 &&\textrm{; $1$ mult.}\\ &> 1 + 1 &&\textrm{; $n>1$}\\ &> 1. &&\textrm{; $1>0$} \end{align*}

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  • $\begingroup$ Woooooow, your explanation in the last section is something. I've never thought it could be something cool and simple like that. But now I feel myself dumb, hope this level of logic will come to me someday) $\endgroup$
    – curioushuman
    Commented Jun 23, 2022 at 19:17
  • $\begingroup$ Also I'm a little confused about that you wrote "... provided that n>1"(the last section as well), but "n" is a natural number, so when "n" is a natural number and > 1, why wouldn't just write "n ≥ 2" ? $\endgroup$
    – curioushuman
    Commented Jun 23, 2022 at 19:17
  • $\begingroup$ And also, thank you very much for your answer! I've upvoted it just now, and if you are curious why I haven't marked your answer as the right one, that's because I don't think that there can be a right answer, because initially the purpose of my question for me was to hear what people think in general about that questions I've asked, so hope you're not offended :) And thank you again !) $\endgroup$
    – curioushuman
    Commented Jun 23, 2022 at 19:21
  • $\begingroup$ While I'm not offended, I must say that you are wrong. There is actually a correct answer, as the formal definition of mathematical proof does exist. My answer is based on such a rigorous notion, only that I've presented it here in a more pedagogical way, to encourage you (and others) to enhance your intuition with more practice and careful reflection. $\endgroup$
    – Leandro Caniglia
    Commented Jun 24, 2022 at 4:32
  • $\begingroup$ Okay, I'm agree, you encouraged me and showed that the proof-making skills are coming with hard work and much experience. Anyway, what about my second comment ? $\endgroup$
    – curioushuman
    Commented Jun 24, 2022 at 12:50

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