I've decided to learn the basics of proofs and here is my first attempt. Could I improve or simplify my proof in any way? Is my formal language correct? Thanks!

Let $n$ be any integer. $$n(n+2) = (n+1)^2 - 1$$ I will prove this identify by induction.

First, check with $n=1$;

$$1 \times 3 = 2^2 + 1 \equiv 3 = 3$$

Inductive step: Assume that the identity is true for n = k;

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

When $n = k + 1$;

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

$$\equiv k+1(k+3) = (k+2)^2 - 1$$

Let $n = k + 1$;

$$n(n+2) = (n+1)^2 - 1$$

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    $\begingroup$ @Danny King: It is common practice among students who want to prove an equality to start with the equality they want to prove, and manipulate both sides until they get something which is true. THIS IS TERRIBLY BAD FORM! It is possible to start with something false, manipulate it, and get something true. Instead, you should either work from one side exclusively until you manage to transform into the other side of the equation, or you should manipulate each side separately until you can get them to look the same. I would take exception to your line right after "when $n=k+1$" for that reason $\endgroup$ Jan 30, 2011 at 19:17
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    $\begingroup$ Another thing to notice is that induction is not the only way to formally prove things, even though introductory courses might make it seem so. Oftentimes induction proofs are opaque while other proofs are simpler and offer more intuition. In this case, induction is used to verify the identity, whereas polynomial manipulation proves it, and also gives us a way of telling ahead of time which similar identities will hold. $\endgroup$ Jan 30, 2011 at 22:20
  • $\begingroup$ Thank you these criticisms are very useful to me. I'll try to apply what you have said in the future. $\endgroup$
    – Danny King
    Jan 31, 2011 at 15:56
  • $\begingroup$ Arturo, regarding your advice on manipulating both sides, does my second proof found at math.stackexchange.com/questions/19716/… also suffer from this? I could see no other way to proceed. Is it allowable in that circumstance? Thanks! $\endgroup$
    – Danny King
    Jan 31, 2011 at 16:20
  • $\begingroup$ It makes absolutely no sense to prove this by induction. $\endgroup$
    – quanta
    May 3, 2011 at 10:53

3 Answers 3


Let $n$ be an integer. $$n(n+2) = n^2 + 2n = n^2 + 2n + 1 - 1 = (n+1)^2 - 1$$


Using $a^2 - b^2 = (a-b)(a+b)$ we have $$(n+1)^2 - 1 = (n+1)^2 - 1^2 = (n+1-1)(n+1+1) = n(n+2).$$


Your proof is correct (except for parentheses missing around $k+1$ somewhere in the induction step), but since you're looking for advice: you could be clearer halfway through the induction step, like: "the l.h.s. yields ..., while the r.h.s. yields ...", making it easier to read. Finally Alo Mestanogullari's comment is spot on; if you'd present this proof more formally, you could use terms like "because of the distributive law..." etc.


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