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I am trying to prove that ($a^n$+$b^n$) is divisible by ($a$+$b$) when $n$ is odd.

I start off by checking for $n$ = $1$, and find that it is true for $n$ = $1$

Now I assume that it is true for $n$ = ($2k$-$1$),

i.e [$a^{2k-1}$+$b^{2k-1}$] is divisible by ($a$+$b$)

Finally, I proceed to checking for $n$ = ($2k$+$1$) :

$=$ {$a^{2k+1}$+$b^{2k+1}$}

$=$ {$a^2$$a^{2k-1}$+$b^2$$b^{2k-1}$}

Now I don’t understand how to prove that {$a^2$$a^{2k-1}$+$b^2$$b^{2k-1}$} is divisible by ($a$+$b$) based on my assumption earlier.

Appreciate any help

4d_
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2 Answers2

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Hint:

$$a^{2k+1}+b^{2k+1}=a^2(a^{2k-1}+b^{2k-1})-b^{2k-1}(a^2-b^2)$$

J. W. Tanner
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For the induction step in this proof to work, we need to check two base cases: $n = 1$ is obvious as you said, and for $n=3$:

$$ a+b\ |\ (a+b)(a^2-ab+b^2) = a^3+b^3$$

Since we proved two base cases, now we can assume it is true for $n=2k-3$ and $n=2k-1$, that means:

$$a+b\ |\ a^{2k-3}+b^{2k-3}\ \wedge\ a+b\ |\ a^{2k-1}+b^{2k-1}$$

Therefore:

$$a+b\ |\ (a^2+b^2)(a^{2k-1}+b^{2k-1})-a^2b^2(a^{2k-3}+b^{2k-3}) = a^{2k+1}+b^{2k+1}$$

we get that

$$a+b \ | \ a^{2k+1}+b^{2k+1}$$

LHF
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  • Need? One could, but there is no real need. –  Feb 19 '20 at 20:12
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    For his proof, it's needed, because of the exponent of $2k-3$ – David P Feb 19 '20 at 20:14
  • @lepidon, your idea also uses $P(1)$. The point is $P(2k-1)$ by itself does not imply $P(2k+1)$ – LHF Feb 19 '20 at 20:14
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    @Atticus What I am saying is that it is not a need of the problem to have to check $P(3)$, as a result doesn't need to be proven by this specific argument. –  Feb 19 '20 at 20:17
  • @Atticus Can we assume that it is true for $n$ = $2k$-$1$ and $n$ = $2k$-$3$ both? I thought we generally assume that it is true for $n$ = $k$ (which is $2k$-$1$ in this case), and then we prove for $n$ = $k$+$1$ (which is $2k$+$1$ in this case) ? – 4d_ Feb 19 '20 at 20:17
  • @πtimese: cf. strong induction – J. W. Tanner Feb 19 '20 at 20:20
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    @DavidPeterson A need of a particular line of argument is not necessarily a need of all possible proofs of a given result. What I said is that the statement "For the induction step to work, we need to check two base cases." can be read as saying that it is a requirement of the result, but this is not true for this result. –  Feb 19 '20 at 20:20
  • @lepidon, using my idea needs it. What I wrote is true up to the semantics. The fact that your idea does not need $P(3)$ specifically, I congratulate you for it. – LHF Feb 19 '20 at 20:21
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    @πtimese, for the induction step, we can assume as many cases are true as we've proven in the base step. – LHF Feb 19 '20 at 20:22
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    I know. But precision is crucial in mathematics. Since the phrase that you used means something different than what you intended and since the literal meaning is false, it is good to clarify such that someone who is learning induction doesn't get the wrong idea. –  Feb 19 '20 at 20:24
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    @lepidon, I edited. I hope it is sufficiently precise now. – LHF Feb 19 '20 at 20:25
  • @Atticus So if I prove three base cases, I can assume three different cases to be true, and use them for my proof later? – 4d_ Feb 19 '20 at 20:33
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    @πtimese, Yes, that's it. – LHF Feb 19 '20 at 20:35
  • @Atticus thanks for clarifying. Can you briefly explain why you proved two base cases? You said using your idea needs it. I still don't quite understand why you had to prove two base cases for the induction step to work. Is it because you later assumed two different cases to be true? – 4d_ Feb 19 '20 at 20:45
  • @πtimese, I added some detail. I hope it's more clear now. – LHF Feb 19 '20 at 20:49
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    @Atticus So basically you proved two base cases because you wanted to assume two cases to be true, to use them in the proof? – 4d_ Feb 19 '20 at 20:52
  • @πtimese, Yes, thats pretty much it. But it's not that I wanted, for the identity I used, I needed both the cases $2k-1$ and $2k-3$, otherwise it wouldn't work. – LHF Feb 19 '20 at 20:54
  • @Atticus I understand it now, thanks a lot. One last question : When we assume two or more cases to be true, they must be for successive values of $n$, right? For example, in this case, the two cases assumed to be true were for $n$ = $2k$-$3$ and $n$ = $2k$-$1$ ? – 4d_ Feb 19 '20 at 20:58
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    @πtimese, Yes, that's true. – LHF Feb 19 '20 at 20:59
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    @Atticus thanks again. It was really helpful – 4d_ Feb 19 '20 at 21:01