I am trying to find all integers $m$ such that $m$ is relatively prime to 30, and $m=6x^2+5y^2$ for some integers $x,y$. Note that we must have: $y$ is odd, $(y,3)=1=(x,5)$. Using these conditions, I was able to show that $m=11$ or $29$ mod $30$. So I was first trying to handle the case $m=11$ mod 30. Note that $11=6\cdot1^2+5\cdot1^2$, but $41$ is not of the form $6x^2+5y^2$. So I was wondering, for which $n$ do we have $30n+11=6x^2+5y^2$? Also, for such $n$, can we obtain $x$ and $y$ satisfying $30n+11=6x^2+5y^2$?
Asked
Active
Viewed 124 times
2
-
There is usually no requirement that $x \neq y$ in these representations so your representation of $11$ is good. – Ross Millikan Jan 16 '23 at 02:57
-
1mistake to use mod 30, correct is mod 120. Legendre symbol $$ (-120|41) = (3|41) = (41|3) = (2|3) = -1. $$ – Will Jagy Jan 16 '23 at 03:17
2 Answers
4
The values to look for are mod 120, specifically $11, 29, 59, 101 \pmod{120}.$ There is a fairly clean theory about primes that are represented by a binary form. Since the class number $h(\mathbb Z[\sqrt{-30}]) = 4$ but there is one form in each genus, the primes $p = 5 x^2 + 6 y^2 $ are described precisely by congruences.
Other than the primes $2,3,5,$ the primes such that $(-120|p) = 1$ are represented as such:
$$ \begin{array}{ccccccc} 1. & 1,&31,&49,&79, & \pmod{120}& :: \; \; x^2 + 30 y^2 \\ 2. & 17,&23,&47,&113, & \pmod{120}& :: \; \; 2x^2 + 15 y^2 \\ 3. & 13,&37,&43,&67, & \pmod{120}& :: \; \; 3x^2 + 10 y^2 \\ 6. & 11,&29,&59,&101, & \pmod{120}& :: \; \; 5x^2 + 6 y^2 \\ \end{array} $$
Products of such primes are represented by one of the four forms, specified by Gauss composition. For example, $11 \cdot 29 = 319 = 17^2 + 30 \cdot 1^2 = 7^2 + 30 \cdot 3^2.$ Suppose we call the forms 1,2,3,6, where the last one is for $6 y^2 + 5 x^2.$ For a product of such "good" primes, write each one with a code 1,2,3,6. Then multiply, table $1^2 = 2^2 = 3^2 = 6^2 = 1.$ Next $2 \cdot 3 = 6$ and $2 \cdot 6 = 3.$ Finally $3 \cdot 6 = 2.$ Well, here is a $2 \cdot 3 = 6.$ As in $17 \cdot 13 = 221 = 5 \cdot 5^2 + 6 \cdot 4^2 = 5 \cdot 1^2 + 6 \cdot 6^2 $
all values mod 120
0 5 6 11 14 20 21 24 26 29
30 35 36 44 45 50 51 54 56 59
60 66 69 74 75 80 84 86 90 96
99 101 104 110 114 116
prime to 120: 11, 29, 59, 101
primes
5 11 29 59 101 131 149 179 251 269
389 419 461 491 509 659 701 821 941 971
1019 1061 1091 1109 1181 1229 1259 1301 1451 1499
1571 1619 1709 1811 1901 1931 1949 1979 2069 2099
2141 2309 2339 2381 2411 2459 2531 2549 2579 2621
2699 2741 2789 2819 2861 2909 2939 3011 3221 3251
3299 3371 3389 3461 3491 3539 3581 3659 3701 3779
3821 3851 3989 4019 4091 4139 4211 4229 4259 4349
4421 4451 4691 4931 5021 5051 5099 5171 5189 5261
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
form 1 x^2 + 30 y^2
0 1 4 9 16 25 30 31 34 36
39 46 49 55 64 66 79 81 94 100
111 120 121 124 129 130 136 144 145 151
156 169 174 184 196 199 201 220 225 226
241 255 256 264 270 271 274 279 286 289
295 306 316 319 324 334 345 351 354 361
370 376 391 400 409 414 430 439 441 444
466 471 480 481 484 489 495 496 505 514
516 520 526 529 544 559 561 576 580 594
601 604 606 624 625 631 649 655 670 676
form 2 2 x^2 + 15 y^2
0 2 8 15 17 18 23 32 33 47
50 60 62 65 68 72 78 87 92 98
110 113 128 132 135 137 143 153 158 162
167 177 185 188 200 207 215 222 233 240
242 248 257 258 260 263 272 288 290 297
302 303 312 335 338 348 353 368 375 377
383 392 393 398 402 407 423 425 440 447
450 452 465 473 482 503 510 512 527 528
537 540 542 548 558 572 575 578 585 590
593 612 617 632 638 647 648 663 668 690
form 3 3 x^2 + 10 y^2
0 3 10 12 13 22 27 37 40 43
48 52 58 67 75 85 88 90 93 102
108 115 117 118 138 147 148 157 160 163
165 172 187 192 198 202 208 232 235 237
243 250 253 262 268 277 282 283 298 300
307 310 325 333 340 352 358 360 363 372
373 387 390 397 403 408 432 435 442 453
460 468 472 490 493 502 507 517 522 523
538 547 550 552 565 588 592 597 598 603
613 628 637 640 643 652 660 667 675 678
form 6 5 x^2 + 6 y^2
0 5 6 11 20 24 26 29 44 45
51 54 59 69 74 80 86 96 99 101
104 116 125 131 134 141 149 150 155 170
176 179 180 186 195 204 216 221 230 234
236 245 251 261 269 275 276 294 296 299
314 320 326 330 339 341 344 374 384 389
395 396 404 405 411 416 419 429 459 461
464 470 474 486 491 500 501 506 509 524
531 536 539 554 555 564 566 596 600 605
611 614 620 621 629 645 650 659 666 680
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
Oscar Lanzi
- 39,403
Will Jagy
- 139,541
-
In this answer I explore the factorization characteristics on class 4 domains when the class group is specifically Klein-four. Would this result be connected with that property of $\mathbb Z>\sqrt{-30}]$? – Oscar Lanzi Jan 16 '23 at 04:47
-
@OscarLanzi Hi. The unusually clean setup is due to having one class per genus, for positive forms this is Euler's Idoneal Numbers. https://en.wikipedia.org/wiki/Idoneal_number I will see if I can come up with one class per genus but eight genera, this might require indefinite forms. Take a few minute at least. Quicker than expected, idoneal number 105, class number eight, all forms square to the identity, 420: < 1, 0, 105> 420: < 2, 2, 53> 420: < 3, 0, 35> 420: < 5, 0, 21> 420: < 6, 6, 19> 420: < 7, 0, 15> 420: < 10, 10, 13> 420: < 11, 8, 11> – Will Jagy Jan 16 '23 at 04:59
-
Alright, 16 genera, one class per genus, Discriminant -5460 h : 16 Squares : 1 cubes : 16 Fourths : 1 5460: < 1, 0, 1365> 5460: < 2, 2, 683> 5460: < 3, 0, 455> 5460: < 5, 0, 273> 5460: < 6, 6, 229> 5460: < 7, 0, 195> 5460: < 10, 10, 139> 5460: < 13, 0, 105> 5460: < 14, 14, 101> 5460: < 15, 0, 91> 5460: < 21, 0, 65> 5460: < 26, 26, 59> 5460: < 30, 30, 53> 5460: < 35, 0, 39> 5460: < 37, 4, 37> 5460: < 42, 42, 43> – Will Jagy Jan 16 '23 at 05:09
-
OEIS A000926 reports that the class group for the relevant discriminants is indeed $(\mathbb Z/2\mathbb Z)^n$, which is the Klein four-group for $\mathbb Z[\sqrt{-30}] (n=2)$. – Oscar Lanzi Jan 16 '23 at 15:01
-
I made your class number rendering more precise (giving the ring, not just the discriminant which may apply to other rings). Feel free to roll back if you want. – Oscar Lanzi Jan 22 '23 at 22:56
0
if n>0, then left hand side is positive, and 30n+11 chooses an ellipse from right-hand side which contains an ellipse for each value in $\{ (x,y) | f(n,x,y)=g(n,x,y), (n,x,y) \in (Z,Z,Z) \}$ where $f(n,x,y)=30n+1$ and $g(n,x,y)=6x²+5y²$. After you realize that it's just an ellipse for positive values of the left hand side, then the next step is to realize that the integers are a grid that needs to intersect the ellipse.
tp1
- 660