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

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.

• $x^2+6y^2-xy=47 \implies (2 x - y)^2 + 23 y^2 = 188$. In pari/gp thue(x^2+23,188) with output [], i.e. no solutions. Apr 4, 2020 at 6:59

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 !

• This is the second unreadable multiline display I've seen from you today. Your answer would be more readable in an align environment. Apr 3, 2020 at 23:26
• The comment was deleted, but it said that my answer was "isomorphic" to yours. If you think it does not provide anything more useful than others posted, please let me know.
– robjohn
Apr 6, 2020 at 6:13
• @robjohn: Yeah, I just double check that we make appropriate comments or edits and follow the MSE's policies. Apr 6, 2020 at 6:41
• @DeepSea: the question was if you don't think my answer adds anything useful. I am considering removing my answer.
– robjohn
Apr 6, 2020 at 7:27
• @robjohn: I looked at your answer and it uses the substitution and complex variables which is not bad at all honestly. Your answer reminded me about using number fields to solve Diophantine Equations.I would keep it. It pointed to a new direction when we solve Diophantine Equations. Apr 6, 2020 at 7:36

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.

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

  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]

?


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

• (+1) the only answer that doesn't simply check the necessary cases.
– robjohn
Apr 8, 2020 at 3:53

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.

• The comment was deleted, but it said that my answer was "isomorphic" to yours. If you think it does not provide anything more useful than others posted, please let me know.
– robjohn
Apr 6, 2020 at 6:13
• @robjohn I don't think our answers are "isomorphic" (I hadn't seen yours yet). But they are rather similar; yours seems like a more detailed version of mine. Apr 6, 2020 at 8:29

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 $$(2x-y)^2$$.

• I was adding more content about getting the quadratic form when I was interrupted. If it still seems too close, I have no problem deleting.
– robjohn
Apr 6, 2020 at 0:03
• Will Jagy's answer is the only one that doesn't check cases.
– robjohn
Apr 8, 2020 at 3:52

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.