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The question is this:

"We assume that $a$,$b$ are positive numbers(they might not be integers) but $a+b$ and $a*b$ are integers, prove that for every $n$ $\in$$\mathbb{N}$ the numbers $a^n + b^n$ are integers.

Hint(Use Binominal Theorem).

I've tried to solve it but I couldn't. I could use some help

S.C.B.
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Shadi
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2 Answers2

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You don't need the Binomial Theorem. Use induction.

Note that $a^n+b^n$ holds for $n=0,1$ from the conditions provided. Now asssume it is true for $n=m$, $n=m+1$. Then using Newton's Identity $$a^{m+2}+b^{m+2}=(a+b)(a^{m+1}+b^{m+1})-ab(a^m+b^m)$$ We find it holds for $m+2$ as well.

Thus, our proof is done.

EDIT

Upon second consideration, one can prove this using the Binomial Theorem by using strong induction. Assume that the proposition holds true for all $n<m$.

If $m$ is even, $$a^m+b^m=(a+b)^m-\sum_{i=1}^{\frac{m-2}{2}} \binom{m}{i} a^ib^i(a^{m-i}+b^{m-i})-\binom{m}{\frac{m}{2}}a^{\frac{m}{2}}b^{\frac{m}{2}} \in \mathbb{Z}$$ By the inductive hypothesis. If $m$ is odd, $$a^m+b^m=(a+b)^m-\sum_{i=1}^{\frac{m-1}{2}} \binom{m}{i} a^ib^i(a^{m-i}+b^{m-i})\in \mathbb{Z}$$ From the inductive hypothesis. We are done.

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S.C.B.
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  • Thank you so much for the answer,But i have one more question about your answer, why did you subctract it by ab(a^m + b^m) ? – Shadi Jan 26 '17 at 15:38
  • @Shadi Your welcome. :) I suggest you accept an answer to take this off the unanswered portion. You can do that by clicking the checkmark. – S.C.B. Jan 26 '17 at 15:39
  • @Shadi To make $a^m+b^m$. Without it we have unnecessary terms. See the link I left in the answer. – S.C.B. Jan 26 '17 at 15:40
  • It actually looks challenging to prove this with the Binomial theorem... –  Jan 26 '17 at 15:44
  • @YvesDaoust this question is made up for 2 parts, the first part is to write the biniominal thoerim equation and the other part is to solve this with a hint that this is the method that we should use,but seeing an answer like this made it look much easier... i couldn't solve it with binomial theorem after 1 and a half hour of trying and i'm supposed to solve it in 10mins – Shadi Jan 26 '17 at 15:46
  • @YvesDaoust See my edit. :) – S.C.B. Jan 26 '17 at 15:57
  • Yep, it seems that you need both the BT and induction. So the hint seems a little sadistic :) –  Jan 26 '17 at 17:27
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$$(a^n+b^n)(a+b)=a^{n+1}+b^{n+1}+ab(a^{n-1}+b^{n-1})$$

$$\iff a^{n+1}+b^{n+1}=(a^n+b^n)(a+b)-ab(a^{n-1}+b^{n-1})$$

Now use Strong Induction.

See : Proving that $\frac{\phi^{400}+1}{\phi^{200}}$ is an integer.

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lab bhattacharjee
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  • It seems we posted the same answer, about $21$ seconds apart. – S.C.B. Jan 26 '17 at 15:29
  • @S.C.B., Yours is without the derivation of the Proposition and without mentioning of Strong Induction? – lab bhattacharjee Jan 26 '17 at 15:31
  • See also : http://math.stackexchange.com/questions/517440/whats-the-difference-between-simple-induction-and-strong-induction and http://math.stackexchange.com/questions/1184541/what-exactly-is-the-difference-between-weak-and-strong-induction – lab bhattacharjee Jan 26 '17 at 15:32
  • What do you mean? We don't need to derive it for the proposition, as it is provided in the condition. I mentioned induction. – S.C.B. Jan 26 '17 at 15:34
  • We don't need strong induction for this, but I find that difference isn't very big. – S.C.B. Jan 26 '17 at 15:35
  • @S.C.B., By proposition, I meant $$a^{m+2}+b^{m+2}=(a+b)(a^{m+1}+b^{m+1})-ab(a^m+b^m)$$ Is it weak induction? – lab bhattacharjee Jan 26 '17 at 15:36