Prove that there do not exist integers m and n such that

5m^2 − 6mn + 7n^2 = 2011.

I don't know how to start please help me

Guest Jul 28, 2020

#1**+2 **

**Prove that there do not exist integers m and n such that**

**\(5m^2 − 6mn + 7n^2 = 2011\).**

\(\begin{array}{|rclrcl|} \hline \mathbf{5m^2 - 6mn + 7n^2} &=& \mathbf{2011} \quad | \quad *5 \\\\ 25m^2 - 30mn + 35n^2 &=& 2011*5 \\ 25m^2 - 30mn + 9n^2+ 26n^2 &=& 2011*5 \\ (5m-3n)^2+ 26n^2 &=& 2011*5 \quad | \quad \mathbf{2011*5} \equiv \mathbf{6 \pmod{13}} \\ \hline \\ (5m-3n)^2+ 26n^2 &\equiv& 6 \pmod{13} \quad | \quad \mathbf{26n^2\pmod{13}} \equiv \mathbf{0} \\ (5m-3n)^2 +0 &\equiv& 6 \pmod{13} \\ (5m-3n)^2 &\equiv& 6 \pmod{13} \quad | \quad x= 5m-2n \\ \mathbf{x^2} &\equiv& \mathbf{6 \pmod{13}} \\ \hline \end{array}\)

**The Quadratic Reciprocity Law with Legendre symbol:**

Let \(\mathbf{p}\) be an odd prime. The integer \(\mathbf{a}\), prime to \(\mathbf{p}\), is said to

be a \(\mathbf{\text{quadratic residue}}\) or \(\mathbf{\text{nonresidue}}\) of \(\mathbf{p}\) according as the congruence

\(x^2 \equiv a\pmod{p}\)

is of is not solvable. The Legendre symbol \(\left(\dfrac{a}{p}\right)\) is defined to be \(+1\) or \(-1\)

according as \(\mathbf{a}\) is a quadratic residue or nonresidure of \(\mathbf{p}\).

\(\begin{array}{|rcll|} \hline \mathbf{x^2} &\equiv& \mathbf{6 \pmod{13}} \quad | \quad a = 6,\ p = 13 \\\\ \left(\dfrac{a}{p}\right) &=& \left(\dfrac{6}{13}\right) \\ \hline \\ \mathbf{\left(\dfrac{6}{13}\right)} &=& \left(\dfrac{2*3}{13}\right) \\\\ &=& \left(\dfrac{2}{13}\right)\left(\dfrac{3}{13}\right) \quad &| \quad \left(\dfrac{3}{13}\right)=\left(\dfrac{13}{3}\right) \\\\ &=& \left(\dfrac{2}{13}\right)\left(\dfrac{13}{3}\right) \\\\ &=& \left(\dfrac{2}{13}\right)\left(\dfrac{13\pmod{3}}{3}\right) \\\\ &=& \left(\dfrac{2}{13}\right)\left(\dfrac{1}{3}\right) \quad &| \quad \left(\dfrac{1}{3}\right) = 1 \\\\ &=& \left(\dfrac{2}{13}\right)*1 \\\\ &=& \left(\dfrac{2}{13}\right) \\\\ &=& \left(-1\right)^{ \frac{13^2-1}{8} } \\\\ &=& \left(-1\right)^{21} \\\\ &=& \mathbf{-1} \\ \hline \end{array}\)

\(\mathbf{\left(\dfrac{6}{13}\right)}=-1\), there do not exist integers m and n such that

\(5m^2 - 6mn + 7n^2 = 2011\).

heureka Jul 28, 2020