Timeline for Can repunits be perfect cubes?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 4, 2013 at 14:28 | comment | added | KConrad | For $y^2 = x^3 + k$ with $k \not= 0$: in ${\mathbf Z}_2$ use $(x,k+1)$ s.t. $x^3 = (k+1)^2 - k$, in ${\mathbf Z}_3$ use $(1-k,y)$ s.t. $y^2 = (1-k)^3+k$, in ${\mathbf Z}_5$ use $(x,y)$ s.t. $x^3 + k \equiv 1 \bmod 5$. In ${\mathbf Z}_7$, use $(0,y)$ if $k \equiv 1, 2, 4 \bmod 7$, $(1,y)$ if $k \equiv 0, 3 \bmod 7$, $(-1,y)$ if $k \equiv 5 \bmod 7$, and $(x,0)$ if $k \equiv 6 \bmod 7$. For $p \geq 11$, let $N_p$ be number of mod $p$ solns, so $N_p = p + S_p$ with $S_p=0$ if $p|k$ and $|S_p| \leq 2\sqrt{p}$ by Hasse if $(p,k)=1$. Then $N_p \geq 4$; lift mod $p$ soln w/ $y \not\equiv 0$ mod $p$. | |
| Jun 4, 2013 at 0:42 | comment | added | KConrad | so there's a mod $p$ solution $(x,y)$ with $y \not\equiv 0 \bmod p$, and such a mod $p$ solution can be lifted $p$-adically. | |
| Jun 4, 2013 at 0:41 | comment | added | KConrad | @Noam: You're right that there's no congruence argument: $y^2 = x^3 + 7$ is solvable in the $p$-adic integers for all primes $p$. First let's treat $p \leq 7$. If $p = 2$ use $(x,0)$ s.t. $x^3 = -7$. If $p = 3$ use $(0,y)$ s.t. $y^2 = 7$. If $p = 5$ use $(-1,y)$ s.t. $y^2 = 6$. If $p = 7$ use $(1,y)$ s.t. $y^2 = 8$. If $p > 7$, let $N_p$ be the number of $(x,y) \bmod p$ s.t. $y^2 \equiv x^3 + 7 \bmod p$. Then $N_p = p + \sum (\frac{x^3+7}{p})$, w/ sum being over $x \bmod p$. Call the sum $S_p$, so $|S_p| \leq 2\sqrt{p}$ by Weil's bound. Then $N_p \geq p - 2\sqrt{p} > 3$ (since $p>9$) [contd] | |
| May 31, 2013 at 15:55 | comment | added | Noam D. Elkies | @Wangt Fei: thanks for Accepting my answer, and for pointing out the typo, which I'll fix in the next edit when I also acknowledge other comments and give references to V.Lebesgue's work. For the descent, the idea is that if $n=2k>0$ and $R_n$ is a $q$-th power then the same is true of $R_k$, because $R_k$ is a unitary divisor of $R_n$ (i.e. $R_n/R_k$ is an integer relatively prime to $R_k$). If $k$ is also even then $R_{k/2}$ is a $q$-th power, etc., and eventually we'd reach a solution of $R_n = m^q$ with $n$ odd, to which we can apply the other argument. | |
| May 31, 2013 at 13:56 | comment | added | Wangt Fei | I find that if n is even and q≡3(mod 4) and q>3, this method seems not work. If we use descent, k(=n/2) would not be an even number and the equation is not the same as the former. | |
| May 30, 2013 at 14:24 | vote | accept | Wangt Fei | ||
| May 30, 2013 at 14:24 | comment | added | Wangt Fei | It's a great proof for a large range of this problem! Thank you very much. The ideal you already gave is elementary but fascinating. However, I want to point out a Clerical error in the paragraph "The same approach deals with..." that "9m^q+1=10^n-1" should be "9m^q=10^n-1". I did not know much aboult two-square theorem in the composite case. Is it because "4x+3" has a "4y+3" type prime factor then (4x-1)(4x+3) can not be decomposed into two squares? | |
| May 29, 2013 at 14:02 | comment | added | Noam D. Elkies | @Joe Silverman: yes, that's a clear example of the same idea, and much earlier than 1980... Thanks. Here there are no solutions at all, but there still doesn't seem to be an easy congruence argument. (mwrank says there isn't even a rational solution, but doing descents on such a curve isn't elementary either!) $$ $$ @Gerry Myerson: As I already mentioned, the fact that a decimal repunit other than $1$ cannot be a fifth power is an immediate consequence of the fact that all fifth powers are congruent to $0$, $\pm 1$, or $\pm 7 \bmod 25$. | |
| May 29, 2013 at 6:01 | comment | added | Gerry Myerson | The Zentralblatt review of Muller's paper, by Olaf Ninnemann, says more about the Rotkiewicz paper, Elem Math 42 (1987) 76, Zbl 0703.11016; it's not quite so elementary, relying on Ljunggren 1943. | |
| May 29, 2013 at 5:58 | comment | added | Gerry Myerson | The review by Nikos Tzanakis of Tom Muller, Note on the Diophantine equation $1+2p+(2p)^2+\cdots+(2p)^n=y^p$, Elem Math 60 (2005) 148-149, MR2188007 (2006h:11038), says that the paper uses nothing beyond Fermat's Little Theorem, and that a direct consequence is that no repunit can be a fifth power. Tzanakis says this extends a result of A Rotkiewicz that repunits can't be squares or cubes, but no reference is given. Strangely, when I go to the webpage that should have the Muller paper, it has all the papers from that issue except Muller's. | |
| May 29, 2013 at 3:16 | comment | added | Joe Silverman | Hi Noam. Your "strange step" reminds me of V.A. Lebsgue's solution of y2=x3+7, although it's not as elaborate. First note x must be odd, else 7 would be a square mod 8. Then add 1 to get $$y^2+1=x^3+8=(x+2)(x^2-2x+4)=(x+2)((x-1)^2+3).$$ The last factor is 3 mod 4, so is divisible by a prime $q\equiv3\pmod4$. Then $y^2+1\equiv0\pmod{q}$, contradicting the fact that $-1$ is not a square mod $q$. So similar to your solution, he uses only the computation of $(-1|q)$. I don't know the exact reference, but I think that it was late 19th century, so it's not a new idea. | |
| May 29, 2013 at 1:11 | history | answered | Noam D. Elkies | CC BY-SA 3.0 |