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A student of mine offered the following proof of the irrationality of $\sqrt{3}$: Suppose $(a/b)^2 = 3$ with $a,b$ having no common factor. Since $a^2=3b^2$, an easy parity argument (using the fact that $a$ and $b$ cannot both be even) shows that $a$ and $b$ must both be odd. Writing $a=2m+1$ and $b=2n+1$ we obtain $(2m+1)^2=3(2n+1)^2$, so $4m^2+4m+1=12n^2+12n+3$. Subtracting $1$ from both sides and dividing by $2,$ we get $2m^2+2m=6n^2+6n+1$, which is impossible since the LHS is even and the RHS is odd.

Has anyone seen this before?

YCor
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James Propp
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    I don't know of any formal source, but a colleague recently asked me the same question after a student submitted this exact solution as homework. – David Conlon Jul 01 '21 at 19:49
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    In the title I mentally substituted $\zeta(3)$ for sqrt(3) and was prepared to be astonished... – Sam Hopkins Jul 01 '21 at 19:51
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    This is more or less equivalent to looking at $a^2=3b^2$ mod 4. Abstractly, you could say that this proof deduces the irreducibility of $x^2-3$ over the rationals by showing it over the $2$-adics. That really makes it qualitatively different from the usual descent argument, which works $3$-adically. – Achim Krause Jul 01 '21 at 19:55
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    Note that it generalizes very quickly to show that if $k \equiv 3$ (mod 4), then $k^{\frac{1}{2}}$ is irrational. – JoshuaZ Jul 01 '21 at 20:02
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    @David Conlon: Hmm. I gave this as a homework problem too. Maybe someone with a Chegg or Coursehero account should check to see if the proof is there... – James Propp Jul 01 '21 at 20:13
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    This has been studied for a while now. The argument fails with 17 rather than 3 and there is a natural question of whether just using parity arguments one can prove the irrationality of $\sqrt 17$. I wrote a thread about it last year. https://twitter.com/AndresECaicedo1/status/1302697810446483456?s=20 – Andrés E. Caicedo Jul 01 '21 at 21:34
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    At first I was confused why it was interesting, but now I see it uses only parity arguments, which is indeed interesting. (Though, while novel, is certainly not nearly as illuminating as pretty much any other proof!) – exfret Jul 01 '21 at 21:43
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    "Has anyone seen this before?" Yes, 2400 years ago in Theaetetus. – Olivier Jul 01 '21 at 22:19
  • @SamHopkins Me too, my brain automatically replaced sqrt by zeta. – efs Jul 02 '21 at 02:11
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    Searching with Approach0 led me to this answer: Prove that the square root of 3 is irrational. (I posted this comment on MO, but it was lost because of the migration.) – Martin Sleziak Jul 02 '21 at 09:27
  • I have seen (somewhere) the comparable proof that $\sqrt{2}$ is irrational by noting that if $a^{2} = 2b^{2}$ for coprime integers $a,b$, then $a^{2} \equiv b^{2} \equiv 1$ (mod $3$), since $a$,$b$ are not both divisible by $3$ (so neither is). – Geoff Robinson Jul 02 '21 at 16:00
  • It comes up in music when explaining why repeated just tempered perfect fifths (frequency ratio of $\frac{3}{2}$) can never be an exact number of octaves (frequency ratio a power of $2$). – badjohn Jul 08 '21 at 07:51

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