Timeline for When is the product $(1+1)(1+4)…(1+n^2)$ a perfect square?
Current License: CC BY-SA 3.0
8 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| S Dec 31, 2013 at 4:41 | history | suggested | José Hdz. Stgo. | CC BY-SA 3.0 |
added a link
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| Dec 31, 2013 at 4:11 | review | Suggested edits | |||
| S Dec 31, 2013 at 4:41 | |||||
| Dec 28, 2013 at 0:17 | vote | accept | Ken W. Smith | ||
| Dec 28, 2013 at 0:10 | comment | added | Lucia | @JohnJiang: If $p$ divides $a^2+1$ and $b^2+1$ then $p$ divides $|a-b|$ or $(a+b)$. If $p>2n$ and $a$ and $b$ are distinct numbers below $n$, this is impossible. So a prime larger than $2n$ that divides the product, only divides to exponent $1$. | |
| Dec 28, 2013 at 0:04 | comment | added | John Jiang | Why does the fact that the largest prime factor exceeds 2n shows the product is not a perfect square? Presumably it should be pretty obvious but I just don't see it. | |
| Dec 27, 2013 at 15:33 | comment | added | Alvin | In fact, the first to improve Chebushev's result was Nagell (reference [4] from Cilleruelo paper), who showed that the largest prime factor of the product $\prod f(j)$ is $\ge n\log n.$ Ciruello's proof goes along Nigell's one by making the estimate precise and using computer search for small $n.$ | |
| Dec 27, 2013 at 15:02 | history | answered | Lucia | CC BY-SA 3.0 |