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Let $P(x)$ be a cubic polynomial with integer coefficients. Does there exist a constant $c$ such that at least one of the following values $P(0),P(1),...,P(c)$ is not a square?

It is known that the number of integral points of $y^2 = P(x)$ is bounded by a constant dependent on the coefficients of $P(x)$ and I am wondering whether in this particular case, it is possible to find a constant independent of the coefficients. Any links to existing literature or maybe an explanation of why such a problem would be impossible to solve with current tools would be of a great help too.

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    $\begingroup$ Just to bound $c$ from below: the polynomial $24x^3-135x^2+192x$ has square values at $0,1,2,3,4$. $\endgroup$ – Ilya Bogdanov Feb 25 '14 at 13:05
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    $\begingroup$ Let me just point out that you can't hope to find a bound for $c$ by local methods. Indeed, for any $N>0$, we can choose integers $a_1,a_2,a_3$ that are $p$-adically small enough, so that $a_3x^3+a_2x^2+a_1x+1$ will be a square in $\mathbb{Z}_p$ for all $0\leq x\leq N$. $\endgroup$ – RP_ Feb 25 '14 at 13:29
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    $\begingroup$ Another one square from 0 to 4 is $2x^3 - 10x^2 + 12x$ $\endgroup$ – joro Feb 25 '14 at 14:47
  • $\begingroup$ Related question: mathoverflow.net/questions/50661/… $\endgroup$ – Lucia Feb 25 '14 at 15:48
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    $\begingroup$ @René: Actually, see Theorem C in Silverman's 1983 paper "Heights and the specialization map for families of abelian varieties." (An elementary special case which suffices here is proven in Masser's Appendix C to Zannier's book "Some problems of unlikely intersections in arithmetic and geometry.") Already it implies that for a given $N$, there are just finitely many elliptic curves over $\mathbb{Q}$ on which, for some Weierstrass equation $y^2 = P(x)$, the points with $x$-coordinates $0,1,\ldots,N$ (and one choice for the square roots in the $y$-coordinates) are dependent. $\endgroup$ – Vesselin Dimitrov Feb 25 '14 at 18:51
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I show here that the existence of such a constant $c$ (and much more) follows from previous conjectures.

Namely, a special case of the Caporaso-Harris-Mazur conjecture asserts that there is a constant $N$ such that: for any squarefree degree-$6$ polynomial $f(x)\in\mathbf{Q}[x]$, there are at most $N$ rational numbers $b$ for which $f(b)$ is a square (in $\mathbf{Q}$). So if $P(x)$ is a cubic polynomial in $\mathbf{Q}[x]$ for which $P(x)P(x+1)$ is squarefree, then there are at most $N$ rational numbers $b$ for which both $P(b)$ and $P(b+1)$ are squares, whence you can take $N+1$ to be your constant $c$. If $P(x)$ is squarefree but $P(x)P(x+1)$ is not, then one can use the above argument with $P(x)P(x+r)$ for some $r\in\{2,3,4\}$, as at least one of these will be squarefree. If $P(x)$ is not squarefree then we can divide it by a square factor in order to reduce to the analogous question for degree-one polynomials, where it's easy to exhibit such a value $c$.

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[EDITED to fix typos, show smoothness of $V_c$, and extend the exhaustive-search result]

Solutions of $P(i) = y_i^2$ ($0 \leq i \leq c$) with $P(X) = \sum_{j=0}^3 a_j X^j$ are parametrized up to scaling by a threefold $V_c$ in projective space ${\bf P}^c$ that's the complete intersection of $c-3$ quadrics $$ y_i^2 - 4 y_{i+1}^2 + 6 y_{i+2}^2 - 4 y_{i+3}^2 + y_{i+4}^2 = 0 \phantom{0\infty} (0 \leq i \leq c-4), $$ minus the points on the hyperplane $y_0^2 - 3 y_1^2 + 3 y_2^2 - y_3^2 = 0$ where the leading coefficient $a_3$ vanishes. We expect plenty of points for $c < 7$, a sparse but still infinite set of points for $c = 7$, and only finitely many points for $c > 7$ except possibly on a proper subvariety. This last part is a special case of the Bombieri-Lang conjecture, and if we assume this conjecture for $V_8$ then we can probably use the forgetful maps $V_c \rightarrow V_8$ for $c>8$ to prove (more directly than using Caporaso-Harris-Mazur as Michael Zieve proposed) that some $V_c$ has no rational points except on the hyperplane $a_3=0$.

[EDIT René's comment raises the question of whether this complete intersection is smooth. The answer is yes in characteristic zero. Any linear combination of the differentials of the quadrics $y_i^2 - 4 y_{i+1}^2 + 6 y_{i+2}^2 - 4 y_{i+3}^2 + y_{i-4}^2$ has the form $(a_0 y_0, a_1 y_1, \ldots, a_c y_c)$ with $\sum_{m=0}^c Q(m) a_m = 0$ for any polynomial $Q(m)$ of degree at most $3$. Therefore in any nonzero combination at least $5$ of the $a_m$ do not vanish. Therefore at a singularity at least $5$ of the $y_i$ must be zero. But this is impossible because $y_i^2$ are valuees of a cubic polynomial at distinct points $i=0,1,\ldots,c$, and at most $3$ of those can be zero unless the polynomial vanishes identically.]

An exhaustive search for rational points on $V_7$ with $a_3 \neq 0$ and $0 \leq y_2,y_3,y_4,y_5 < 1024$ finds only the following $22$ [EDIT extended from $1024 = 2^{10}$ to $1536 = 3 \cdot 2^9$, and found eight more solutions, for a total of $30$], up to the symmetry $y_i \leftrightarrow y_{7-i}$:

13 7 1 1 5 7 7 1
53 21 7 29 45 53 49 3
1586 847 24 73 610 861 868 221
139 23 31 115 173 209 217 181
1061 577 35 73 469 721 883 935
31 52 47 34 35 64 107 158
821 433 49 127 355 479 473 79
139 83 71 97 125 139 127 41
359 19 79 299 439 509 481 229
163 124 107 110 121 128 121 82
169 157 119 67 55 131 233 349
368 247 134 35 76 163 242 311
826 481 164 1 286 451 544 539
323 223 167 167 197 223 223 167
595 379 187 17 109 205 251 217
973 109 239 817 1195 1387 1319 679
497 323 247 283 353 397 377 203
676 499 266 121 416 799 1226 1691
34 369 332 185 138 419 784 1203
2258 1259 356 235 838 1183 1216 521
1393 927 449 229 705 1273 1873 2499
2836 1597 470 223 1016 1471 1562 925
1373 889 475 211 317 497 581 475
1179 728 581 750 977 1124 1113 794
2027 1315 673 199 499 917 1255 1487
1027 1064 749 50 343 1252 2239 3334
2573 1629 791 65 507 923 1121 969
961 896 817 802 925 1196 1579 2042
1297 1082 827 608 613 922 1393 1948
1144 1135 854 413 388 1099 1970 2951

none of these lifts to a rational point on $V_8$ (neither $P(-1)$ nor $P(8)$ is a square); possibly $c=8$ is already small enough th make a solution impossible.

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  • $\begingroup$ Very nice answer. By the way, the minus in one of the indices in your displayed formula should be a plus. (I deleted my previous comment, whose validity rested on the assumption that the $V_c$ are smooth. I thought I had an easy proof of this, but it was wrong.) $\endgroup$ – RP_ Feb 27 '14 at 16:26
  • $\begingroup$ Thanks! I fixed the typo you noted (and also un-overloaded the index $i$), and gave a proof of the smoothness of $V_c$ (and took the opportunity to extend the exhaustive search a bit). $\endgroup$ – Noam D. Elkies Mar 1 '14 at 5:57
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Lower bound for $c$ is 6 $7$.

$$ f(x) = -5096*x^3 + 70161*x^2 - 232960*x + 262144 $$ is square for $ x \in [0,6]$.

This is specialization of the identity which is square for $x \in [0,5]$:

$$ P_5 = \left(2 b_{0} + \frac{75}{8}\right) x^{3} + \left(-15 b_{0} - \frac{4311}{64}\right) x^{2} + \left(\frac{115}{4} b_{0} + 120\right) x + b_{0}^{2}$$

For $0 \le x \le 7$

$g(x)=4*x^3 - 36*x^2 + 80*x + 1$

If I remember correctly, the related question for quadratic is conjecturally bounded by about $8$.

Added

The $P_5$ identity gave infinite family of $8$ solutions arising from genus $1$ curve (to avoid scaling require the content to be square free).

I believe absolute bound on $c$ follows from the conjectures about absolute bound of number of rational points on curves of genus $ > 1$.

Since there are no four consecutive squares in arithmetic progression, the cubic $f(x)$ is square-free.

This leads to the curve $[f(x)=y^2,f(x+1)=z^2]$ which I expect to be genus $4$ in general (and possibly always).

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  • $\begingroup$ A smaller example that works for $0 \leq x \leq 5$ is $-4x^3+24x^2-20x+1$. $\endgroup$ – RP_ Feb 25 '14 at 15:36
  • $\begingroup$ @René what is the expected rank of the elliptic curves? $\endgroup$ – joro Feb 25 '14 at 15:38
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    $\begingroup$ $4*x^3 - 36*x^2 + 80*x + 1$ works for [0,7] $\endgroup$ – joro Feb 25 '14 at 15:42
  • $\begingroup$ I don't know what to expect. My cubic gives an elliptic curve with rank $2$, which is slightly disappointing of course... $\endgroup$ – RP_ Feb 25 '14 at 15:44
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    $\begingroup$ The $[0,7]$ example comes from the rank-$3$ curve of minimal conductor $5077$; I learned about it from Dick Gross in connection with the paper $$ $$ Joe P. Buhler, Benedict H. Gross, and Don B. Zagier: On the Conjecture of Birch and Swinnerton-Dyer for an Elliptic Curve of Rank 3 Mathematics of Computation 44 #170 (April 1985), 473-481 $$ $$ $-$ see Table 1 on page 476: people.mpim-bonn.mpg.de/zagier/files/doi/10.2307/2007967/… $\endgroup$ – Noam D. Elkies Feb 26 '14 at 6:18

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