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If m is a power of $3$, n is a power of $3$, prove that $m+n$ is never a power of $3$.

This is the question that I was given, unfortunately, my teacher doesn't like to teach and I' left reading a $1000$ page book by myself answering these random questions that aren't covered in our book.

Edit: I think I really need help defining what is meant by a power of $3$. When they say $m$ is a power of $3$ does that mean $m=3\enspace\text{or}\enspace A^m = \enspace\text{or}\enspace A^3$ ?
if so $A^m + B^n$ with $m$ & $n$ being powers of $3 == A^3 + B^3$ and therefore $m+n = 3+3 = 6$ and a power of $6$ is not a power of $3$ ?

Kumar
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Karl
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    Try a proof by contradiction. Assume that $m=3^i, n=3^j, m+n=3^k$. What can you conclude? – Kevin Long Sep 25 '19 at 15:52
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    Additional hint to what @Kevin Long said: $3^4 + 3^9 = 3^4(1 + 3^5).$ – Dave L. Renfro Sep 25 '19 at 15:53
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    For any $m$, $m$ is a power of $a$ if and only if there exists an $n$ such that $m=a^n$. Thus, $9$ is a power of $3$ because there exists a number, that is, the number $2$, such that $9=3^2$ – RyRy the Fly Guy Sep 25 '19 at 16:16
  • Unrelated to the question, but please don't put unnecessary remarks regarding your negative opinions towards your instructor in the question body. It is distracting, totally unnecessary, and very disrespectful. – YiFan Tey Oct 06 '19 at 01:44
  • @Karl, if you are satisfied with your answer, then please click the green check to close the post. Thanks! – RyRy the Fly Guy Nov 04 '19 at 04:34

2 Answers2

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Given $m=3^x$ and $n=3^y$, assume $m+n=3^x+3^y=3^z$. Note that any power of $3$ is an odd number. An odd number plus an odd number cannot equal an odd number. Hence, you have a proof by contradiction. I leave it to you to show that powers of $3$ are odd.

RyRy the Fly Guy
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  • okay so maybe I understand....if 3^m + 3^n = 3^a then a could only be a negative number such as 9 or 27 thus a could never = m+n? – Karl Sep 25 '19 at 16:17
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    @Karl In General, the following result holds true: Any power of an even number is even and any power of odd is odd. Moreover, the proof is simple just by the use of induction. – Kumar Sep 25 '19 at 16:24
  • @Karl I guess I could interpret your last comment. What you say is not correct. It is not necessary that $a=m+n$. It could be a different number altogether. For Eg: $2^5+2^5=2^6$, here, as per your last comment, $m=n=5$ and $a=6$ but $6\neq 5+5=10$. – Kumar Sep 25 '19 at 16:33
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Powers of 3 and cubes are different things. Given an exponent $\alpha$ that is a positive integer, $3^\alpha$ is a power of 3. If you flip that, however, $\alpha^3$, you have a cube, and the only way that's also a power of 3 is if $\alpha = 1$ or 3.

The first few powers of 3 are: 1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907. These are listed in Sloane's OEIS A000244.

The first few cubes are 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331, 1728, 2197, 2744, 3375, 4096, 4913, 5832, 6859, 8000, 9261, 10648, 12167, 13824, 15625, 17576, 19683, 21952, 24389. These are listed in Sloane's A000578.

The only numbers in common to both lists are 1 and 27.

So if $m$ is a power of 3 and $n$ is also a power of 3, meaning that $m = 3^\alpha$ and $n = 3^\beta$ (it doesn't matter if $\alpha = \beta$, as long as they're both positive integers), then $m + n = 3^\alpha + 3^\beta$.

Not much help there, until you notice, like the other answerer already mentioned, that powers of 3 are odd, but adding up two of them gives an even number.

Now, to prove that two nonzero cubes can never add up to a cube, well, Fermat claimed to have a wonderful proof of that which was just a wee bit too long for the margin.

One more thing. If a number is a power of 3 other than 1, its representation in the ternary numeral system consists of a single digit 1 followed by one or more digits 0. If a number is the sum of two powers of 3, then its representation in ternary is either:

  • A single digit 2 followed by one more digits 0, or
  • A digit 1, followed by one or more digits 0, followed by a 1 and then one or more digits 0, or
  • Two digits 1 with one or more digits 0 in between them.

These numbers are also listed in Sloane's OEIS. The first few are 6, 10, 12, 18, 28, 30, 36, 54, 82, 84, 90.

Robert Soupe
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