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Are there integers $m$ and $n$ such that $$5m^2-6mn+7n^2 = 1985?$$

Taking the equation modulo $3$ gives $n^2-m^2 \equiv 2 \pmod{3}$. Thus, $3 \mid n$ but $3 \nmid m$. How can I use this to find a contradiction?

Martin Sleziak
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user19405892
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5 Answers5

5

The discriminant of the quadratic form is $6^2-4\cdot 5\cdot 7=-2^3\cdot \color{blue}{13}$ and the equation is equivalent to:

$$ (5m-3n)^2+26 n^2 = 9925 \tag{1}$$ that gives: $$ 6\equiv 9925 \equiv (5m-3n)^2 \pmod{\color{blue}{13}}\tag{2} $$ but $6$ is not a quadratic residue $\!\!\pmod{13}$, since $13\equiv 5\pmod{8}$ and $13\equiv 1\pmod{3}$, so: $$ \left(\frac{6}{13}\right)=\left(\frac{-1}{13}\right)\left(\frac{2}{13}\right)\left(\frac{-3}{13}\right)=(+1)\cdot(-1)\cdot(+1)=\color{red}{-1}\tag{3}$$ and there are no integer solutions to the original equation.

For similar problems, quadratic reciprocity and prime factors of the discriminant are the key.

Jack D'Aurizio
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  • I've just tried to answer this question and googled about it: Why 13 being a factor of the discriminant suggests we should look for something $\pmod{13}$? – Red Banana Oct 28 '20 at 23:53
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Not as elegant as Jack D'Aurizio's answer, but here's a way to use the fact that $3\mid n$.

After rewriting the equation as

$$(5m-3n)^2+26n^2=9925$$

let $n=3N$ and $M=5m-3n$, so the equation becomes

$$M^2+234N^2=9925$$

It's easy to see that this requires $0\le N\le6$, and it's straightforward to check that $\sqrt{9925-234N^2}$ is not an integer in that range.

Barry Cipra
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Taking modulo $5$ we have

$$n(2n-m)\equiv 0\pmod 5$$

This gives us two possibilities:

  • If $n$ is a multiple of $5$, write $n=5u$ and $175u^2-30mu+5m^2=1985$, or $35u^2-6mu+m^2=397$. Then

$$m=\frac{6u\pm\sqrt{1588-104u^2}}{2}=3u\pm\sqrt{397-26u^2}$$

  • If $2n\equiv m\pmod 5$, then $m=5u+2n$ and (after some easy calculations) $m^2-14mu-29u^2+397=0$. Now, $$m=\frac{14u\pm\sqrt{312u^2-1588}}2=7u\pm\sqrt{78u^2-397}$$

Both possibilities require $-397$ to be a square modulo $13$, but $-397\equiv -7\pmod{13}$, and $-7$ is not a square modulo $13$.

Thus, there is no solution.

ajotatxe
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Times $7$ (AC method) yields $\ 7\cdot 1985 = N^2\!-6mN\!+\!35m^2 = (N\!-\!3m)^{\large\color{#c00}{\bf 2}}\!+2\cdot \color{#0a0}{13}\,m^2$

So $\,{\rm mod}\ \color{#0a0}{13}\!:\,\ 7\cdot 1985\equiv -2\,$ is a $\rm\color{#c00}{square},\,$ but $\,(-2)^6\equiv 64\equiv -1,\,$ contra Euler's criterion.

Community
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Bill Dubuque
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  • Or $,6$-th powering by little fermat: $\ {\rm mod}\ 13!:\ ({-}2\equiv \color{#c00}{a^2})^{,6}\Rightarrow -1\equiv 1\Rightarrow,13\mid 2,, $ contradiction – Bill Dubuque Jul 20 '16 at 18:01
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If you want to make a boatload more assertions like the one here, here's the surface that $f(m,n) = 5m^2-6mn+7n^2$ describes. I've highlighted the values near $f(m,n)=1985$ (and - just so this an actual answer - none of them are equal to $1985$). Note that $f(-m,-n) = f(m,n)$.

enter image description here

Joffan
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