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Give a direct proof of $8\mid (3^n + 5^n)$ for all odd natural numbers.

I know how to prove this by induction, I am not sure how to go about it using a direct proof.

I would start by saying that $3^n + 5^n = 8k$ for some k in the naturals. But I'm not sure...

Maria
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  • What do you mean by "direct proof" vs. "inductive proof"? For example, is a proof that employs the (trivial) inductive proof $\ (-1)^{2n+1}\equiv -1,$ a "direct" or "inductive" proof of the desired result? – Bill Dubuque Jul 18 '14 at 17:36

5 Answers5

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We have

$$3\equiv -5\mod 8$$ so for $n$ odd we have $$3^n\equiv (-5)^n\mod 8$$ and then the result follows.

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$$5\equiv-3\pmod8$$

Using Congruence Property $\#10$ of this, $$\implies5^{2m+1}\equiv(-3)^{2m+1}\equiv-3^{2m+1}\pmod8$$

See also :

  1. Why $a^n - b^n$ is divisible by $a-b$?

  2. Proof of $a^n+b^n$ divisible by a+b when n is odd

Community
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lab bhattacharjee
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If n is odd then $3^n+5^n=(3+5)(3^{n-1}-3^{n-2}.5+3^{n-3}.5^2-...+5^{n-1})=8.k$ where k is natural number.

Emil Stoyanov
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For even $w,$ $$ 1^w \equiv 3^w \equiv 5^w \equiv 7^w \equiv 1 \pmod 8. $$ I suppose there is a hidden induction in the even part, do it for $w=2,$ then $w=4,$ and so on.

No, wait, take $k$ odd and consider $k^{w/2}$ which is also odd. It is now one of four cases, $k^{w/2} \equiv 1,3,5,7 \pmod 8,$ which all lead to $$k^w = \left( k^{w/2} \right)^2 \equiv 1 \pmod 8.$$ At this stage, one may ask whether the fact that an odd number to any positive integer power is odd comes under the heading of induction. In one way it must, but if already proven in class or the book...

For odd $n,$ take $n=w+1,$ so $$ 1^n \equiv 1, \; 3^n \equiv 3, \; 5^n \equiv 5, \; 7^n \equiv 7 \pmod 8. $$

Will Jagy
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Every power of $3$ is modulo $8$ congruent to either $3$ or $1$. That can be seen quickly as follows: $3^2\equiv 1$, then multiply that by $3$, getting $3$, then multiply that by $3$, getting $1$, then multiply that by $3$, getting $3$, and so on.

Similarly every power of $5$ is modulo $8$ congruent to either $5$ or $1$.

in $3^n+5^n$, you're adding either $3+5$ or $1+1$ depending on whether $n$ is odd or even.

Michael Hardy
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  • Do you really think that is a non-inductive proof? (as the OP seeks) – Bill Dubuque Jul 18 '14 at 18:08
  • @BillDubuque : If one were to fit this proof into a form that is today conventionally considered logically rigorous, it would probably be phrased as a proof by induction. Should one call a proof "inductive" if it was written by a mathematician who had no explicit awareness of the principle of mathematical induction and the proof is one instance of it, not mentioning that it generalizes. Archimedes wrote that if each term in a series $a+b+\cdots+z$ is $1/4$ of the previous term except that the last one is $1/3$ of the previous term, then the sum does not depend on the number of terms. – Michael Hardy Jul 18 '14 at 18:21
  • . . . but Archimedes might never have notice the phenomenon of proof by induction in general. Then there's the question of what would be considered an inductive proof for the purposes of this m.s.e. posting. – Michael Hardy Jul 18 '14 at 18:22
  • I asked because I am curious the way people think about these things, both students and mathematicians. If you use modular arithmetic then you can reduce the induction to the trivial induction that $, 1^n\equiv 1.\ $ Whether or not the resulting proof is "inductive" or "direct" depends on the OP's definition of such. – Bill Dubuque Jul 18 '14 at 18:31