5
$\begingroup$

So, I decided to work with mod $8$ to help develop some intuition on how to generalize the proof. I noticed that taking $a^2$ to clearly be a perfect square, $a^2$ is always congruent to $0,1,-4,4 \mod 8$, at least for a good finite set of cases. Also, I played around with solutions to $3^m + 3^n + 1 = b$ and seen again for a finite set of cases with $m,n$ varying these were always, $b$ is congruent to $-1,5,-5,3$. Now, I see that well only perfect squares produce a set $\{0,1,-4,4\} \mod 8$ and none of the cases I tried to $m,n$ gave me any of the numbers in this set. And now I'm stuck.

$\endgroup$
  • $\begingroup$ Note that $$4\equiv -4\bmod 8$$ and $$-5\equiv 3\bmod 8$$ $\endgroup$ – Zev Chonoles Jun 14 '13 at 3:31
  • $\begingroup$ I'm sorry, I'm not following where this is going. $\endgroup$ – Mr.Fry Jun 14 '13 at 3:36
  • $\begingroup$ I wasn't trying to give you a hint on the problem (you're actually basically done already), I was just pointing out that it is redundant to list $4\bmod 8$ and $-4\bmod 8$ separately, and likewise, $-5\bmod 8$ and $3\bmod 8$ separately. $\endgroup$ – Zev Chonoles Jun 14 '13 at 3:37
  • $\begingroup$ Ah, yeah I see what you mean. Sorry about that. I was just pondering whether I can end this knowing that from a finite set of m,n cases we can see a pattern of solutions that do not match those of perfect squares. $\endgroup$ – Mr.Fry Jun 14 '13 at 3:41
1
$\begingroup$

The simple modulo 8 approach in the question works fine, you just have to extend "try some random samples" on each side to actual proofs.

You can prove that $3^n \bmod 8$ is always either $1$ or $3$, by induction on $n$. From this you can see that $3^n+3^m+1$ is equivalent modulo 8 to either one of $1+1+1=3$ or $3+1+1=1+3+1=5$ or $3+3+1=7$.

On the other hand all perfect squares are either $0$, $1$, or $4$ modulo 8 (which you can see by squaring $0,1,2,\dots,7$ and taking each of them modulo 8).

Since $\{3,5,7\}$ and $\{0,1,4\}$ don't have any possibilities in common, the desired result follows.

$\endgroup$
  • $\begingroup$ I can do this problem now. This is from my first course I took in abstract algebra over this past summer. $\endgroup$ – Mr.Fry Feb 24 '14 at 4:33
  • $\begingroup$ @Hey123: Nevertheless it seems worthwhile to have the straightforward solution on record ... $\endgroup$ – Henning Makholm Feb 24 '14 at 11:12
3
$\begingroup$

It's easy to see that $3^m + 3^n + 1$ is odd. Suppose that $3^m + 3^n + 1 = 4k^2 + 4k + 1$, then $3^m + 3^n \equiv 0 \pmod 8$. We can suppose $m \geq n$ by symmetry, then $3^n(3^{m-n} + 1) \equiv 0 \pmod 8$ and so $3^{m-n} \equiv -1 \pmod 8$. This is a contradiction because just $1$ and $3$ are residues of powers of $3$ modulo $8$.

$\endgroup$
  • $\begingroup$ Can you just explain this a bit more to me. It seems like you used 4k^2+4k+1 because perfect squares can always be expressed this way. What if this wasn't common knowledge? And also can you particularly clarify the second line? Thanks a lot though! $\endgroup$ – Mr.Fry Jun 14 '13 at 4:12
  • 1
    $\begingroup$ As $3^m + 3^n + 1$ is odd and we are assuming $3^m + 3^n + 1$ is a perfect square, it is a perfect square of a odd number, for example, $(2k+1)^2$. But $(2k+1)^2 = 4k^2 + 4k + 1 = 4k(k+1) + 1$. One of $k$ or $k+1$ is even, then $4k(k+1) \equiv 0 \pmod 8$ and so $3^m + 3^n + 1 = 4k^2 + 4k + 1 \equiv 1 \pmod 8 \Rightarrow 3^m + 3^n \equiv 0 \pmod 8 \Rightarrow 3^n(3^{m-n} + 1) \equiv 0 \pmod 8$, but $gcd(3,8) = 1$, then $3^{m-n} + 1 \equiv 0 \pmod 8$. Is it ok now? $\endgroup$ – Savio Jun 14 '13 at 4:30
  • $\begingroup$ Yes, thanks alot. $\endgroup$ – Mr.Fry Jun 14 '13 at 4:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.