Is $441$ the only square of the form $\frac{397\cdot 10^n-1}{9}$?
Can it be proven?
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10$\begingroup$ unmotivated number theory questions tend to be poorly received, you will want to provide a motivation for your question to increase the chances of a productive answer. $\endgroup$– Carlo BeenakkerCommented Feb 20, 2021 at 18:21
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10$\begingroup$ I wish people wouldn't be so quick to close questions like this. If they had thought about the problem posed then they would have realised that it was non trivial. As a professional number theorist, I actually like questions like this: simple to state yet require difficult tools to solve $\endgroup$– Daniel LoughranCommented Feb 20, 2021 at 19:31
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7$\begingroup$ Also as a number theorist I have no problem with unmotivated number theory questions. $\endgroup$– Daniel LoughranCommented Feb 20, 2021 at 19:31
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31$\begingroup$ I respectfully disagree about whether this question is trivial or not. Note that $397 \cdot 10^{n} - 1 \equiv 3 \pmod{4}$ if $n \geq 2$ and hence it cannot be a square. $\endgroup$– Jeremy RouseCommented Feb 20, 2021 at 22:25
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3$\begingroup$ Ha, nice. I do wonder whether those who voted to close noticed this easy argument. Though for more general problems like this I guess using the Mordell equation is the way to go. $\endgroup$– Daniel LoughranCommented Feb 21, 2021 at 6:38
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1 Answer
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If $\frac{397\cdot 10^n - 1}9$ is a square then so is $397\cdot 10^n - 1$. Let $y^2=397\cdot 10^n - 1$. Denoting $x:=10^{\lfloor n/3\rfloor}$, we get that $$y^2 = 397\cdot 10^r\cdot x^3 - 1$$ or $$(397\cdot 10^{r}y)^2 = (397\cdot 10^r\cdot x)^3 - 397^2\cdot 10^{2r}$$ where $r:=n\bmod 3\in\{0,1,2\}$. These are Mordell equations with many known solutions, which in general can be solved by finding integral points on the elliptic curve.
I've solved this equation, and confirm your conjecture.
answered Feb 20, 2021 at 18:29
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2$\begingroup$ The following code snippet for Sage verifies that the only integral points on the corresponding Mordell curves are $(3970,\pm 250110)$, which corresponds to the $441$ which OP asked for. (oops, editor eats the line breaks. Not sure how to fix it) for r in range(3): N = 397*10^r E = EllipticCurve([0,-N^2]) print(r,E.integral_points())` $\endgroup$– WojowuCommented Feb 20, 2021 at 18:58
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$\begingroup$ @Wojowu: Yes, I've got the same. $\endgroup$ Commented Feb 20, 2021 at 19:01