Which integer from 1 to 20 does the quadratic form $<7,11>=7x^2+11y^2$ over $\mathbb{Q}$ represent? This is an exercise from chapter 1 of Lam's book, Introduction to Quadratic forms over fields. All I could do was to prove that it represents 7, 11, 2, 8 and 18. What should I do to solve this? Can I do this without using some difficult facts from number theory?


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    $\begingroup$ Welcome to Math.Stackexchange. A few tips: Post the question in the body as well as in the title of the post. Please add a little more context, for example explaining what you mean by <7,11>. $\endgroup$ – mrf Nov 3 '15 at 10:50
  • $\begingroup$ Thanks for tips mrf! $\endgroup$ – Absol Nov 3 '15 at 12:16

If $7(x/z)^2+11(y/z)^2=1$ then $7x^2+11y^2=z^2,$ and we may assume $\gcd(x,y,z)=1.$ First if $z$ is even we must have $x,y$ odd by the gcd condition, but then mod $8$ each of $x^2,y^2$ is $1$ so the left side is $7+11=18=2$ mod 8, however even squares are only $0,4$ mod $8.$

So $z$ is odd, making $z^2=1$ mod $8$. One of $x,y$ is odd and the other even. So the left side is either $7+11*e^2=7+3e^2$ (with $e$ even) mod $8$ where $e^2$ is $0$ or $4,$ making the left side $7$ or $3$ mod $8$ neither of which works.

Or the left side is $7e^2+11=3,7$ mod $8$ and again neither work.

Similar case work mod $8$ might dispose of some other possibilities, or maybe use another mod. Beyond that I don't have the expertise.

Added: I wrote a simple program which, for a given $d,$ looks for solutions to $7x^2+11y^2=dz^2$ mod $8$ for which $\gcd(x,y,z)=1.$ This would go with representing a given $d$ from $1$ to $20$ as a rational sum of squares with coefficients $7,11,$ as done above for $d=1.$ Unfortunately it only eliminated $d=1,5,6,9,13,14,17$ from the desired $d$ range. So at this point either try to show the others can rise by some concrete examples, or else maybe work modulo something other than $8.$

  • $\begingroup$ Eight numbers from 1 to 20 now remain, for which mod 8 argument doesn't rule them out, and they are not listed as formable by OP. These are 3,4,10,12,15,16,19,20. $\endgroup$ – coffeemath Nov 3 '15 at 17:37
  • $\begingroup$ the easiest thing to remember is Legendre's therem, math.stackexchange.com/questions/27471/… which ought to be stated for squarefree coefficients $a,b,c$ and therefore pairwise coprime. In brief, consideration $\pmod 7$ and $\pmod {11}$ finishes things. It is an exercise on page 185 in Lam $\endgroup$ – Will Jagy Nov 3 '15 at 18:40
  • $\begingroup$ also see math.stackexchange.com/questions/89138/… Legendre's theorem is on pages 80-82 in Cassels, I like his clear statement that one must check every (odd) prime that divides one of the coefficients; it is not until page 82 that he says, if we know the form is indefinite, we can ignore $2,$ that is conditions (ii), (iii) $\endgroup$ – Will Jagy Nov 3 '15 at 18:55
  • $\begingroup$ i.e. they've got to be (quadratic) non-residues $2,6,7,8,10 \pmod {11}$ and, if not divisible by 7, residues $1,2,4 \pmod 7.$ $\endgroup$ – Will Jagy Nov 3 '15 at 21:17
  • $\begingroup$ Thanks for answers! $\endgroup$ – Absol Nov 7 '15 at 0:44

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