Looking for proof of no solution to 4-variable quadratic diophantine equation

Question

Show there are no integers $a,b,c,d$ such that $$\begin{cases}1=9ac+3ad+3bc-16bd \\ 1=3ad+3bc+2bd \end{cases}$$

Motivation: The ideal $I=(3,1+\sqrt{-17})$ in $R=\mathbb{Z}[\sqrt{-17}]$ has the property that that $\omega=1+\sqrt{-17} \in I^2$ but is not a product of two elements of $I$. Writing out the claim (that $\omega$ is a product of two elements of $I$) as a diophantine equation (and noting the integral basis of $I$) gives the equation.

Mild progress: Subtracting the equations gives $0=9(ac-2bd)$ so that $ac=2bd$ and so we get an equivalent system: $$\begin{cases}0=ac-2bd \\ 1=3ad+3bc+2bd \end{cases}$$

Another method: The norm $N(a+b\sqrt{-17}) = a^2+17b^2$ is multiplicative, and the minimum value of $N$ on $I\setminus \{0\}$ is $9$, so the minimum value of a nonzero product of two elements of $I$ is $81$, while the value on $\omega$ is only $18$. However, I'm not familiar with low-tech methods of finding the minimum nonzero norm of a set. In particular how does one show: $$\min\left(\{ 3a^2+2ab+6b^2 : a,b\in\mathbb{Z}; (a,b) \neq 0\}\right) = 3$$ or at least that it is strictly greater than $1$? [ I've shown it is at least equal to 1. ]

Answer 1

We have $3a^2+2ab+6b^2=(a+b)^2+2a^2+5b^2$, which is as a sum of non-negative squares at least $5$, if not $b=0$. Hence we may assume that $b=0$. Then obviously $3a^2\ge 3$. The general estimates can be found in the nice answer of Will.

Answer 2

$$ 4A (A x^2 + B x y + C y^2) = 4 A^2 x^2 + 4 AB x y + B^2 y^2 + (4AC-B^2) y^2= (2Ax+By)^2 + (4AC-B^2) y^2 $$

$$ A x^2 + B x y + C y^2 \geq \left( \frac{4 A C - B^2}{4C}\right) x^2 $$ $$ A x^2 + B x y + C y^2 \geq \left( \frac{4 A C - B^2}{4A}\right) y^2 $$ when $A,C > 0$ and $4AC - B^2 > 0.$

$$ 3 x^2 + 2 x y + 6 y^2 \geq \left( \frac{68}{24}\right) x^2 $$ $$ 3 x^2 + 2 x y + 6 y^2 \geq \left( \frac{68}{12}\right) y^2 $$

$68/24 \approx 2.833333,$ so when $x$ is nonzero and integral,................

If the form $A x^2 + B x y + C y^2$ is positive and reduced, meaning $0 < A \leq C$ and $-A < B \leq A,$ then the minimum of the form for integral $(x,y) \neq (0,0)$ is simply $A$ itself.

Indefinite forms are very different; that would be real quadratic fields..There is an entirely successful approach, due to Lagrange and perhaps Gauss, that gives the "minimum" without needing high decimal accuracy in a continued fraction or any computer memory or pattern matching at all. I have answered many MSE questions with that, I'm not sure anyone understood the advantage over continued fractions; the approaches are mathematically equivalent but not computer implementation equivalent.

The previous indefinite forms were mostly Pell, where the minimum is automatically 1. So, here is an example where i did not know the minimum, I randomly picked coefficients. $-4$ appears at steps 80,81.

    =========================================

        jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./indefCycle
        Input three coefficients a b c for indef f(x,y)= a x^2 + b x y + c y^2 
        173 567 -1020

          0  form            173         567       -1020  delta      0
          1  form          -1020        -567         173  delta      1
          2  form            173         913        -280


                  -1          -1
                   0          -1

        To Return  
                  -1           1
                   0          -1

        0  form   173 913 -280   delta  -3
        1  form   -280 767 392   delta  2
        2  form   392 801 -246   delta  -3
        3  form   -246 675 581   delta  1
        4  form   581 487 -340   delta  -2
        5  form   -340 873 195   delta  4
        6  form   195 687 -712   delta  -1
        7  form   -712 737 170   delta  5
        8  form   170 963 -147   delta  -6
        9  form   -147 801 656   delta  1
        10  form   656 511 -292   delta  -2
        11  form   -292 657 510   delta  1
        12  form   510 363 -439   delta  -1
        13  form   -439 515 434   delta  1
        14  form   434 353 -520   delta  -1
        15  form   -520 687 267   delta  3
        16  form   267 915 -178   delta  -5
        17  form   -178 865 392   delta  2
        18  form   392 703 -340   delta  -2
        19  form   -340 657 438   delta  1
        20  form   438 219 -559   delta  -1
        21  form   -559 899 98   delta  9
        22  form   98 865 -712   delta  -1
        23  form   -712 559 251   delta  3
        24  form   251 947 -130   delta  -7
        25  form   -130 873 510   delta  1
        26  form   510 147 -493   delta  -1
        27  form   -493 839 164   delta  5
        28  form   164 801 -588   delta  -1
        29  form   -588 375 377   delta  1
        30  form   377 379 -586   delta  -1
        31  form   -586 793 170   delta  5
        32  form   170 907 -301   delta  -3
        33  form   -301 899 182   delta  5
        34  form   182 921 -246   delta  -3
        35  form   -246 555 731   delta  1
        36  form   731 907 -70   delta  -13
        37  form   -70 913 692   delta  1
        38  form   692 471 -291   delta  -2
        39  form   -291 693 470   delta  1
        40  form   470 247 -514   delta  -1
        41  form   -514 781 203   delta  4
        42  form   203 843 -390   delta  -2
        43  form   -390 717 329   delta  2
        44  form   329 599 -508   delta  -1
        45  form   -508 417 420   delta  1
        46  form   420 423 -505   delta  -1
        47  form   -505 587 338   delta  2
        48  form   338 765 -327   delta  -2
        49  form   -327 543 560   delta  1
        50  form   560 577 -310   delta  -2
        51  form   -310 663 474   delta  1
        52  form   474 285 -499   delta  -1
        53  form   -499 713 260   delta  3
        54  form   260 847 -298   delta  -3
        55  form   -298 941 119   delta  8
        56  form   119 963 -210   delta  -4
        57  form   -210 717 611   delta  1
        58  form   611 505 -316   delta  -2
        59  form   -316 759 357   delta  2
        60  form   357 669 -406   delta  -2
        61  form   -406 955 71   delta  13
        62  form   71 891 -822   delta  -1
        63  form   -822 753 140   delta  6
        64  form   140 927 -300   delta  -3
        65  form   -300 873 221   delta  4
        66  form   221 895 -256   delta  -3
        67  form   -256 641 602   delta  1
        68  form   602 563 -295   delta  -2
        69  form   -295 617 548   delta  1
        70  form   548 479 -364   delta  -2
        71  form   -364 977 50   delta  19
        72  form   50 923 -877   delta  -1
        73  form   -877 831 96   delta  9
        74  form   96 897 -580   delta  -1
        75  form   -580 263 413   delta  1
        76  form   413 563 -430   delta  -1
        77  form   -430 297 546   delta  1
        78  form   546 795 -181   delta  -4
        79  form   -181 653 830   delta  1
        80  form   830 1007 -4   delta  -252
        81  form   -4 1009 578   delta  1
        82  form   578 147 -435   delta  -1
        83  form   -435 723 290   delta  2
        84  form   290 437 -721   delta  -1
        85  form   -721 1005 6   delta  168
        86  form   6 1011 -217   delta  -4
        87  form   -217 725 578   delta  1
        88  form   578 431 -364   delta  -1
        89  form   -364 297 645   delta  1
        90  form   645 993 -16   delta  -62
        91  form   -16 991 707   delta  1
        92  form   707 423 -300   delta  -2
        93  form   -300 777 353   delta  2
        94  form   353 635 -442   delta  -1
        95  form   -442 249 546   delta  1
        96  form   546 843 -145   delta  -6
        97  form   -145 897 384   delta  2
        98  form   384 639 -403   delta  -2
        99  form   -403 973 50   delta  19
        100  form   50 927 -840   delta  -1
        101  form   -840 753 137   delta  6
        102  form   137 891 -426   delta  -2
        103  form   -426 813 215   delta  4
        104  form   215 907 -238   delta  -4
        105  form   -238 997 35   delta  28
        106  form   35 963 -714   delta  -1
        107  form   -714 465 284   delta  2
        108  form   284 671 -508   delta  -1
        109  form   -508 345 447   delta  1
        110  form   447 549 -406   delta  -1
        111  form   -406 263 590   delta  1
        112  form   590 917 -79   delta  -12
        113  form   -79 979 218   delta  4
        114  form   218 765 -507   delta  -1
        115  form   -507 249 476   delta  1
        116  form   476 703 -280   delta  -3
        117  form   -280 977 65   delta  15
        118  form   65 973 -310   delta  -3
        119  form   -310 887 194   delta  4
        120  form   194 665 -754   delta  -1
        121  form   -754 843 105   delta  8
        122  form   105 837 -778   delta  -1
        123  form   -778 719 164   delta  5
        124  form   164 921 -273   delta  -3
        125  form   -273 717 470   delta  1
        126  form   470 223 -520   delta  -1
        127  form   -520 817 173   delta  5
        128  form   173 913 -280


minimum was   4
rep 
177003899612243834005256403731983 
608947905015187166087607559710201 
disc   1027329 dSqrt 1013.572395  M_Ratio  34.32554
Automorph, written on right of Gram matrix:  
3635174031481176012669428632236695796012068290929256206913  20241115434222530087909412012011999770317408825378480215040
12506117750430348947172600993135985572374684738537418132864  69635668286642497335031189942832966475654190639395514908097
=========================================


        =========================================
        jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$

Answer 3

You only have to show it doesn't equal $1$. Solving the equation for $a$ gives $$ b=\frac{-a\pm \sqrt{6-17a^2}}6 $$ If $a\neq 0$, the root is imaginary, so $a=1$. But then, $b$ is not an integer, so this doesn't have any solutions.

Answer 4

Not sure you need all this high power machinery. Here is my analysis

From $ac = 2 bd$ we have $$ a/b = 2d /c ~\text{ (say) } = s/t$$ Then $$ a = u s \\ b=u t \\ d =v s/2 \\ c= v t$$ Substitute in the second and solve for $u$ $$ \frac{2}{\left( 6\,{t}^{2}+2\,s\,t+3\,{s}^{2}\right) \,v}$$ So $$ v = \pm 1 ~\text{ or } \pm 2$$ and $$ 6\,{t}^{2}+2\,s\,t+3\,{s}^{2} = \pm1 ~\text{ or } \pm 2$$

Argue that $6\,{t}^{2}+2\,s\,t+3\,{s}^{2} $ can never be $\pm 1$ or $\pm 2$ by setting it to $\pm 1$ and $\pm 2$ and solving for $t$.

Answer 5

Suppose we want to make $3a^2+2ab+6b^2$ small, where neither $a$ nor $b$ is $0$. It is clearly best to choose $a$ and $b$ of opposite sign. So we want to make $3x^2-2xy+6y^2$ small, with $x$ and $y$ positive. Note that $$3x^2-2xy+6y^2=3(x-\sqrt{2}y)^2+(6\sqrt{2}-2)xy.$$ For positive $x$ and $y$, the right-hand side is greater than $6$.