Quadratic Diophantine equation $x^2+6y^2-xy=47$ has no solutions.

Question

I am trying to show that $x^2 + 6y^2 - xy = 47$ has no integer solutions. I know that the an efficient way is to look at this equation modulo $n$; other equations can be easily be solved this way. I tried this for $n = 2,3,4,5,6$ so far and I still cannot conclude that no solutions exist. Is there an efficient way of knowing what $n$ to try? Can you give some ideas for $n$ not large? Thanks.

Answer 1

I'm afraid this one cannot be settled efficiently by congruences. As it is positive definite, ruling out $47$ (over the integers) is a short finite check.

In general, given a prime larger than 3 that is a quadratic residue mod 23 we can express $$ p = x^2 - xy + 6 y^2 $$ if and only if there are three distinct roots to $$ t^3 - t + 1 \equiv 0 \pmod {23} $$ If this fails, then we get, instead, $$ p = 2 x^2 -xy + 3 y^2 $$ The way I wrote it, take $x=4, y=3$ to get $47$

Now, given a prime (that does not divide the discriminant) it is represented by either a single form, or one form and its "opposite." The three reduced forms of discriminant $-23$ are $$ x^2 + xy + 6 y^2 $$ $$ 2 x^2 + xy + 3 y^2 $$ $$ 2 x^2 - xy + 3 y^2 $$

You have had questions on quadratic number fields. There is a bijection between the ideal class group and the group of quadratic (positive) forms of the discriminant.

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  x^2 + xy + 6 y^2
     23,     59,    101,    167,    173,    211,    223,    271,    307,    317,
    347,    449,    463,    593,    599,    607,    691,    719,    809,    821,
    829,    853,    877,    883,    991,    997,
parisize = 4000000, primelimit = 500000
? p = 59
%1 = 59
? factormod(x^3 - x + 1,p)
%2 = 
[ Mod(1, 59)*x + Mod(4, 59) 1]

[Mod(1, 59)*x + Mod(13, 59) 1]

[Mod(1, 59)*x + Mod(42, 59) 1]

? p = 101
%3 = 101
? factormod(x^3 - x + 1,p)
%4 = 
[Mod(1, 101)*x + Mod(20, 101) 1]

[Mod(1, 101)*x + Mod(89, 101) 1]

[Mod(1, 101)*x + Mod(93, 101) 1]

? p = 167
%5 = 167
? factormod(x^3 - x + 1,p)
%6 = 
[ Mod(1, 167)*x + Mod(73, 167) 1]

[Mod(1, 167)*x + Mod(127, 167) 1]

[Mod(1, 167)*x + Mod(134, 167) 1]

? p = 173
%7 = 173
? factormod(x^3 - x + 1,p)
%8 = 
[ Mod(1, 173)*x + Mod(97, 173) 1]

[Mod(1, 173)*x + Mod(110, 173) 1]

[Mod(1, 173)*x + Mod(139, 173) 1]

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

  2 x^2 + xy + 3 y^2
      2,      3,     13,     29,     31,     41,     47,     71,     73,    127,
    131,    139,    151,    163,    179,    193,    197,    233,    239,    257,
    269,    277,    311,    331,    349,    353,    397,    409,    439,    443,
    461,    487,    491,    499,    509,    541,    547,    577,    587,    601,
    647,    653,    673,    683,    739,    761,    811,    823,    857,    859,
    863,    887,    929,    947,    967,

? 
? p = 13
%9 = 13
? factormod(x^3 - x + 1,p)
%10 = 
[Mod(1, 13)*x^3 + Mod(12, 13)*x + Mod(1, 13) 1]

? p = 29
%11 = 29
? factormod(x^3 - x + 1,p)
%12 = 
[Mod(1, 29)*x^3 + Mod(28, 29)*x + Mod(1, 29) 1]

? p = 31
%13 = 31
? factormod(x^3 - x + 1,p)
%14 = 
[Mod(1, 31)*x^3 + Mod(30, 31)*x + Mod(1, 31) 1]

? p = 41
%15 = 41
? factormod(x^3 - x + 1,p)
%16 = 
[Mod(1, 41)*x^3 + Mod(40, 41)*x + Mod(1, 41) 1]

? p = 47
%17 = 47
? factormod(x^3 - x + 1,p)
%18 = 
[Mod(1, 47)*x^3 + Mod(46, 47)*x + Mod(1, 47) 1]

? p = 71
%19 = 71
? factormod(x^3 - x + 1,p)
%20 = 
[Mod(1, 71)*x^3 + Mod(70, 71)*x + Mod(1, 71) 1]

? 

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Answer 2

Regard it as a equation in $x$, and rewrite: $x^2 - yx + 6y^2 - 47 = 0\implies \triangle = y^2 - 4(6y^2-47) = 188 - 23y^2\ge 0\implies y^2 \le 8\implies |y| = 0,1,2$ . And none of them yield a perfect square for $\triangle$. Thus no integer solutions !

Answer 3

If $(x,y)$ is an integral solution to your equation, then $$4\times47=4x^2 + 24y^2 - 4xy = (2x-y)^2+23y^2,$$ which shows that $y^2\leq8$ and hence $|y|\leq2$. A quick check of the corresponding five quadratics in $x$ yields no integral solutions.

Answer 4

Let $z=x/y$. Since the roots of $z^2-z+6$ are $\frac{1\pm i\sqrt{23}}2$, we get $$ z^2-z+6=\left(z-\frac{1+i\sqrt{23}}2\right)\left(z-\frac{1-i\sqrt{23}}2\right) $$ Multiplying by $y^2$ gives $$ \begin{align} 47 &=x^2-xy+6y^2\\[9pt] &=\left(x-\frac{1+i\sqrt{23}}2\,y\right)\left(x-\frac{1-i\sqrt{23}}2\,y\right)\\ &=\left(x-\frac12\,y\right)^2+\left(\frac{\sqrt{23}}2\,y\right)^2 \end{align} $$ Multiplying by $4$ yields $$ x^2+6y^2-xy=47\iff(2x-y)^2+23y^2=188 $$ Since $23\cdot3^2=207\gt188$, the only choices for $y$ are $\{0,\pm1,\pm2\}$. $$ \begin{array}{r|c} y&188-23y^2\\\hline 0&188\\ \pm1&165\\ \pm2&96 \end{array} $$ None of the numbers in the right column are perfect squares, so none can be $(2x-y)^2$.

Answer 5

The equation $x^2-xy+y^2-47=0$ represents an ellipse and it is easy to verify that for $y\ge3$ and $y\le-3$ there are not real roots for the quadratic resultant.

Consequently it is enough to verify the corresponding values for $x$ when $y$ take the values $y=\pm2,\pm1,0$. We have $$y=-2\Rightarrow x^2+2x-23=0\\y=-1\Rightarrow x^2+x-41=0\\y=0\Rightarrow x^2-47=0\\y=1\Rightarrow x^2-x-41=0\\y=2\Rightarrow x^2-2x-23=0$$ None of these five equations have integer roots. Thus the given diophantine equation has no solution.