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Is the following conjecture obviously false? I have checked on a computer for $n$ up to $200$. Since that is pretty low, and there doesn't seem to be any good upper bound on the minimum value of $t$ needed, I wouldn't be surprised if it was still false.

Let $n > 1$ be squarefree. Then there exists a $t \geq n$ such that all the numbers $$n + t^2 - k^2, \qquad 0 \leq k < t$$ have squarefree part greater than $n$.

Of course I would love to kill this with something like Schinzel's hypothesis H, but that doesn't seem to be enough since the list of polynomials involved grows with the input.

Here is the list of the minimal $t \geq n$ that works for each squarefree $1 < n \leq 203$, organized as ordered triples $(n, t, s)$, where $t$ is the minimum required $t$, and $s$ is the smallest squarefree part of all the $n + t^2 - k^2$. At first I was hopeful that one could choose $t$ close to $n$, but some of the data points actually need very large $t$. Perhaps the thing to do is to parameterize $t = an + b$ with $b < n$, and hopefully observe that $a = 1$ and small $b$ works a lot of the time.

(2, 2, 5)

(3, 6, 14)

(5, 5, 14)

(6, 6, 17)

(7, 8, 22)

(10, 10, 29)

(11, 12, 34)

(13, 13, 38)

(14, 14, 41)

(15, 30, 74)

(17, 30, 19)

(19, 22, 62)

(21, 21, 62)

(22, 25, 71)

(23, 30, 82)

(26, 27, 79)

(29, 33, 94)

(30, 30, 89)

(31, 32, 94)

(33, 78, 41)

(34, 34, 101)

(35, 70, 174)

(37, 85, 206)

(38, 38, 113)

(39, 108, 62)

(41, 41, 122)

(42, 126, 69)

(43, 46, 55)

(46, 46, 137)

(47, 48, 67)

(51, 54, 158)

(53, 57, 166)

(55, 110, 274)

(57, 95, 86)

(58, 62, 181)

(59, 64, 71)

(61, 93, 66)

(62, 67, 78)

(65, 65, 194)

(66, 77, 74)

(67, 74, 214)

(69, 93, 254)

(70, 91, 103)

(71, 122, 314)

(73, 143, 358)

(74, 84, 86)

(77, 135, 106)

(78, 390, 114)

(79, 92, 262)

(82, 95, 271)

(83, 92, 266)

(85, 187, 149)

(86, 86, 257)

(87, 135, 89)

(89, 137, 122)

(91, 148, 130)

(93, 177, 138)

(94, 110, 178)

(95, 120, 334)

(97, 355, 193)

(101, 101, 302)

(102, 2346, 134)

(103, 112, 326)

(105, 2947, 145)

(106, 154, 157)

(107, 176, 458)

(109, 113, 334)

(110, 143, 167)

(111, 252, 219)

(113, 145, 402)

(114, 300, 137)

(115, 320, 146)

(118, 128, 373)

(119, 206, 146)

(122, 150, 421)

(123, 222, 131)

(127, 166, 458)

(129, 207, 146)

(130, 845, 139)

(131, 208, 194)

(133, 427, 206)

(134, 158, 449)

(137, 161, 218)

(138, 462, 237)

(139, 148, 434)

(141, 159, 173)

(142, 191, 163)

(143, 268, 182)

(145, 377, 226)

(146, 162, 181)

(149, 307, 158)

(151, 244, 166)

(154, 322, 173)

(155, 160, 179)

(157, 205, 178)

(158, 225, 166)

(159, 742, 183)

(161, 189, 538)

(163, 338, 838)

(165, 3555, 506)

(166, 188, 541)

(167, 481, 239)

(170, 961, 311)

(173, 235, 642)

(174, 212, 206)

(177, 954, 213)

(178, 202, 197)

(179, 277, 183)

(181, 359, 229)

(182, 226, 233)

(183, 732, 219)

(185, 435, 274)

(186, 186, 557)

(187, 850, 238)

(190, 535, 359)

(191, 216, 211)

(193, 727, 241)

(194, 244, 206)

(195, 880, 211)

(197, 355, 218)

(199, 386, 259)

(201, 471, 218)

(202, 229, 659)

(203, 212, 218)

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  • $\begingroup$ What are the polynomials you are talking about? $\endgroup$ Sep 2, 2020 at 15:40
  • $\begingroup$ @JohnGowers: sorry, I just meant that if instead we only required those numbers to be squarefree when $0 \leq k < N$ for some constant $N$, then maybe the conjecture would follow from Schinzel's hypothesis H, using the polynomials $X^2 + n - k^2$ (of course we would get them to be simultaneously prime, not just squarefree). $\endgroup$
    – babu_babu
    Sep 2, 2020 at 16:49
  • $\begingroup$ For large $n$ , it will be more and more difficult to find such a $t$. Maybe it helps , if you post one solution $t$ for each of the squarefree numbers from $2$ to $200$. It is well possible that the Schinzel-hypothesis (generalization of the Bunyakovsky conjecture) implies this claim. $\endgroup$
    – Peter
    Sep 3, 2020 at 8:37
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    $\begingroup$ @Peter: Thanks! I have posted the smallest possible $t$ for each such $n$, plus the minimum of the squarefree parts of the relevant numbers. Most of them are close to $n$, but some of them are not which is what made me nervous about whether this is true $\endgroup$
    – babu_babu
    Sep 3, 2020 at 22:45
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    $\begingroup$ I arrived at $n=751$, the process more and more slows down. I think, we need restrictions to find the number $t$ more quickly. $\endgroup$
    – Peter
    Sep 4, 2020 at 13:20

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