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# HELP NEEDED!!!

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

5m^2 − 6mn + 7n^2 = 2011.

Jul 28, 2020

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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
$$5m^2 - 6mn + 7n^2 = 2011$$. Jul 28, 2020
edited by heureka  Jul 28, 2020