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Prove that $3a^2-1$ is never a perfect square when $a$ is an integer.

I'm not sure how to go about this proof or what form of an integer to use. I know an integer can be represented using

  • $2k$, $2k+1$, or

  • $3k$, $3k+1$, $3k+2$, or

  • $4k$, $4k+1$, $4k+2$, $4k+3$...

    but how do I know which form to use for this problem?

Community
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Lil
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5 Answers5

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Hint $ $ If false let $\:\!n\:\!$ be the $\rm\color{#0a0}{least}$ natural with $\,\color{#c00}{n^2 =3k-1}\,$ for some $\,k\in \Bbb Z.\,$ Clearly $\,n\neq \color{#90f}{0,1,2}\,$ thus $\,n\!-\!3\,$ is a $\rm\color{#0a0}{smaller}$ such natural by $\:\!(n\!-\!3)^2\! =\!\!\underbrace{\color{#c00}{n^2}-6n+9}_{\textstyle \color{#c00}3(\color{#c00}k-2n+3)\!\color{#c00}{-\!1}\!\!\!\!\!\!\!\!}\!\!,\,$ contra $\,n\,$ is the $\rm\color{#0a0}{least}$.

Note $ $ I avoided mod arithmetic since you wrote it is unknown. Above we used a form of induction known as Fermat's method of infinite descent, or minimal counterexample, or minimal criminal. $ $ If you study modular arithmetic then you can reformulate the above more simply as follows

$$\bmod \color{#c00}3\!:\ n\equiv \color{#90f}{0,1,2}\,\Rightarrow\, \color{#c00}{n^2}\equiv 0,1,1\not\equiv \color{#c00}{{-}1},\ \ {\rm so}\ \ \color{#c00}{n^2+1\neq 3k}\qquad\qquad$$

Bill Dubuque
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    Perusing this thread I was a bit worried that no answer is mentioning the solution based on $-1$ not being a QR modulo $3$. Thank you :-) – Jyrki Lahtonen Sep 02 '15 at 06:42
  • I know this thread is over 8 years old, but: As a beginner in number theory, I had a lot of trouble following this answer. Not only because I'm unsure what "If not" is referring to (it feels like some words were omitted), but also because it's unclear how this line of reasoning follows from the problem statement. My first inclination is to set $3a^2 - 1 = k^2$ and try to induce some sort of contradiction. – Aleksandr Hovhannisyan Nov 10 '22 at 20:45
  • @AleksandrH "If not" and "Deny" etc are common idioms that signal the start of a proof by contradiction, i.e. we start by assuming the negation of the statement being proved, then we deduce a contradiction. Contradictive forms of inductive proofs are often conveniently proved in this minimal criminal form, as already mentioned in my first comment above (posted a few minutes after my answer). Did you miss that? If anything remains unclear then let me and I'll by happy to explain further. – Bill Dubuque Nov 11 '22 at 09:57
  • @Downvoter If something is not clear then please feel welcome to ask questions and I will be happy to elaborate. – Bill Dubuque Nov 11 '22 at 09:58
  • I downvoted this answer because it assumes knowledge of a certain style of proof that most people asking this question (myself included) likely don't have. This is an early question in Burton's elementary number theory. As someone studying this subject for the first time, I generally find it very disheartening when so many of the answers I come across on Math SE assume knowledge of other concepts that my books did not teach. It leads me down a rabbit hole of sub-proofs for prerequisite knowledge until I forget where I started. I found johannesvalks's proof to be more intuitive. – Aleksandr Hovhannisyan Nov 13 '22 at 16:09
  • @AleksandrH The OP gives no source or context for the question (likely it appears in many textbooks). That you have not yet learned the above inductive method does not imply that the same is true for the OP or others at their level. In fact this is a basic method usually taught very early in most quality number theory courses. By downvoting the answer you will lower its visibility to others like yourself that could benefit from this answer helping to close this gap in their knowledge. – Bill Dubuque Nov 13 '22 at 16:44
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Hint: You can use modulo $3$ or modulo $4$. A square number can give only $0$ and $1$ as remainder modulo $4$.

If $a\equiv 0\pmod4$ (meaning $a$ gives remainder $0$) then $3a^2-1\equiv3\cdot0-1\equiv -1\equiv 3 \pmod4$.
If $a\equiv 1\pmod4$, then $3a^2-1\equiv3\cdot1-1=2\pmod4$.

Berci
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    I did not learn mods yet sorry – Lil Feb 27 '14 at 01:21
  • @Lil In other words, a square number can only be represented as $4k$ or $4k+1$ – Mike Feb 27 '14 at 01:25
  • can it be represented as 3k or 3k+1 or 3k+2? – Lil Feb 27 '14 at 01:25
  • Then, first write $a=4k,\ 4k\pm1$ or $a=4k+2$, and prove my first statement. Then write $a^2=4s$ or $a^2=4s+1$ to go on. – Berci Feb 27 '14 at 01:25
  • Well, you mean modulo $3$. It seems even simpler, as $3a^2-1$ is clearly of form $3k+2$ which cannot be a square. – Berci Feb 27 '14 at 01:26
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    It takes fewer steps to work mod $3$. – André Nicolas Feb 27 '14 at 01:27
  • sorry can you guys please break it down for me. So I'm completing a chart for 3k,3k+1,3k+2,3k+3 now to prove 3a^2-1 do I plug that in for k? What is the math that needs to be done to prove the statement? – Lil Feb 27 '14 at 01:29
  • Write these lines $$(3k\pm1)^2=(9k^2\pm6k)\ +1$$ $$(3k)^2=9k^2$$ to get convinced that a square can only give remainder $0$ or $1$ when divided by $3$. Finally, observe that $3a^2-1$ gives remainder $2$ because $$3a^2-1\ =\ (3a^2-3)\ +2,.$$ – Berci Feb 27 '14 at 01:38
  • @Berci There is a small mistake in your second line where it says $3 \cdot 0-2$ instead of $3 \cdot 0-1$. – Sid Feb 27 '14 at 01:38
  • ok. I have a question so I completed the proof but I wanted to try also completing it for mod 4. For mod 4 in order to have a perfect square there would also be a remainder of 0 or 1 when divided by 4 correct? – Lil Feb 27 '14 at 01:46
  • @Lil: Yes, correct: $4k+2$ and $4k$ squares to a $4k'$, and $4k\pm1$ to $4k'+1$. So, for $a^2$ we have two possibilities only: either $a^2$ is of the form $4k$ or $4k+1$. – Berci Feb 27 '14 at 12:05
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Without mods, use parity.

Focus on

$$ 3a^2 - 1 = p^2. $$

So we have two options:

Case 1. $a$ is odd and $p$ is even

$$ 3 ( 2k + 1 )^2 - 1 = 4 \ell^2. $$

Thus

$$ 6 k^2 + 6 k + 1 = 2\ell^2. $$

Which is a contradiction as we get odd = even.

Case 2. $a$ is even and $p$ is odd.

$$ 12 k^2 - 1 = ( 2 \ell + 1 )^2. $$

Thus

$$ 6 k^2 = 2 \ell^2 + 2 \ell + 1. $$

Which is a contradiction as we get even = odd.

So there are no integer solutions.

johannesvalks
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there are two ways to prove that $3a^2-1$ is never a perfect square. first we can prove it by contrapositive and secondly we can prove it by contradiction. So im going to prove by contradiction. Assume that $3a^2-1$ is a perfect square, Then that means $3a^2-1$ can be written in form of $3k$ or $3k+1$.

i.e. $3a^2-1=3k$ or $3a^2-1=3k+1$

If $3a^2-1=3k$ then $3(a^2-k)=1$, which is a contradiction since $3$ does NOT divide $1$.

If $3a^2-1=3k+1$ the $3(a^2-k)=2$, which is also a contradiction since $3$ does NOT divide $2$.

In conclusion $3a^2-1$ is or will never be a perfect square. Q.E.D

Oussama Boussif
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Owen
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0

You can prove it using modular arithmetic:

$n$ is a perfect square $\implies[n\equiv0\pmod4]\vee[n\equiv1\pmod4]$:

  • $k\equiv0\pmod4 \implies k^2\equiv0^2\equiv\color\green0\pmod4$
  • $k\equiv1\pmod4 \implies k^2\equiv1^2\equiv\color\green1\pmod4$
  • $k\equiv2\pmod4 \implies k^2\equiv2^2\equiv\color\green0\pmod4$
  • $k\equiv3\pmod4 \implies k^2\equiv3^2\equiv\color\green1\pmod4$

$a$ is an integer $\implies[3a^2-1\not\equiv0\pmod4]\wedge[3a^2-1\not\equiv1\pmod4]$:

  • $a\equiv0\pmod4 \implies 3a^2-1\equiv3\cdot0^2-1\equiv\color\red3\pmod4$
  • $a\equiv1\pmod4 \implies 3a^2-1\equiv3\cdot1^2-1\equiv\color\red2\pmod4$
  • $a\equiv2\pmod4 \implies 3a^2-1\equiv3\cdot2^2-1\equiv\color\red3\pmod4$
  • $a\equiv3\pmod4 \implies 3a^2-1\equiv3\cdot3^2-1\equiv\color\red2\pmod4$
barak manos
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