Help Me

Here's a rough and ready way to tackle this:

Notice that if b = 2a and c = 3a the left hand side of the first equation becomes 4a² + 6a² + 3a² = 13a²

In other words it is a multiple of 13 and hence satisfies the first equation.

For this situation, if a, b and c are all positive integers less than 13, then a can only be one of 1, 2, 3 or 4.

Trying these in turn we find that only a = 3 (hence b = 6 and c = 9) satisfy the second and third equations.

Hence a + b + c  =  5 mod(13)  

If a, b, and c are positive integers less than 13 such that
\(\large{\begin{align*} 2ab+bc+ca&\equiv 0\pmod{13}\\ ab+2bc+ca&\equiv 6abc\pmod{13}\\ ab+bc+2ca&\equiv 8abc\pmod {13} \end{align*}}\)
then determine the remainder when a+b+c is divided by 13.

\(\begin{array}{|lrcll|} \hline & 2ab+bc+ca &\equiv& 0\pmod{13} \quad |\quad : (abc)\\ (1) & \dfrac{2}{c} + \dfrac{1}{a} + \dfrac{1}{b} &\equiv& 0 \pmod{13} \\ \hline & ab +2bc+ ca &\equiv& 6abc\pmod{13} \quad |\quad : (abc)\\ (2) & \dfrac{1}{c} + \dfrac{2}{a} + \dfrac{1}{b} &\equiv& 6 \pmod{13} \\ \hline & ab + bc+2ca &\equiv& 8abc\pmod{13} \quad |\quad : (abc)\\ (3) & \dfrac{1}{c} + \dfrac{1}{a} + \dfrac{2}{b} &\equiv& 8 \pmod{13} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline 3*(1)-(2)-(3): & 3*\left( \dfrac{2}{c} + \dfrac{1}{a} + \dfrac{1}{b}\right) \\ & -\left( \dfrac{1}{c} + \dfrac{2}{a} + \dfrac{1}{b}\right) \\ & -\left( \dfrac{1}{c} + \dfrac{1}{a} + \dfrac{2}{b}\right) \\ & &\equiv& 3*0 - 6 - 8 \pmod{13} \\ & \dfrac{6}{c} + \dfrac{3}{a} + \dfrac{3}{b} \\ & -\dfrac{1}{c} - \dfrac{2}{a} - \dfrac{1}{b} \\ & -\dfrac{1}{c} - \dfrac{1}{a} - \dfrac{2}{b} \\ & &\equiv& -14 \pmod{13} \\ & \dfrac{4}{c} &\equiv& -14 \pmod{13} \\ & \dfrac{4}{c} &\equiv& 13-14 \pmod{13} \\ & \dfrac{4}{c} &\equiv& -1 \pmod{13} \quad | \quad +1 \\ & \dfrac{4}{c}+1 &\equiv& 0 \pmod{13} \quad | \quad *c \\ & 4+c &\equiv& 0 \pmod{13} \quad | \quad -4 \\ & c &\equiv& -4 \pmod{13} \\ & c &\equiv& 13-4 \pmod{13} \\ & \mathbf{ c } &\equiv& \mathbf{ 9 \pmod{13} } \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline 3*(3)-(2)-(1): & 3*\left( \dfrac{1}{c} + \dfrac{1}{a} + \dfrac{2}{b}\right) \\ & -\left( \dfrac{1}{c} + \dfrac{2}{a} + \dfrac{1}{b}\right) \\ & -\left( \dfrac{2}{c} + \dfrac{1}{a} + \dfrac{1}{b}\right) \\ & &\equiv& 3*8 - 6 - 0 \pmod{13} \\ & \dfrac{3}{c} + \dfrac{3}{a} + \dfrac{6}{b} \\ & -\dfrac{1}{c} - \dfrac{2}{a} - \dfrac{1}{b} \\ & -\dfrac{2}{c} - \dfrac{1}{a} - \dfrac{1}{b} \\ & &\equiv& 18 \pmod{13} \\ & \dfrac{4}{b} &\equiv& 18-13 \pmod{13} \\ & \dfrac{4}{b} &\equiv& 5 \pmod{13} \quad | \quad -5 \\ & \dfrac{4}{b}-5 &\equiv& 0 \pmod{13} \quad | \quad *b \\ & 4-5b &\equiv& 0 \pmod{13} \quad | \quad *(-1) \\ & 5b &\equiv& 0 \pmod{13} \quad | \quad +4 \\ & 5b &\equiv& 4 \pmod{13} \quad | \quad : 5 \\ & b &\equiv& 4*\dfrac{1}{5} \pmod{13} \\ & && \boxed{\dfrac{1}{5} \pmod{13} \\ \equiv 5^{\varphi(13)-1}\pmod{13} \\ \equiv 5^{12-1}\pmod{13} \\ \equiv 5^{11}\pmod{13} \\ \equiv 48828125\pmod{13} \\ \equiv 8\pmod{13} } \\ & b &\equiv& 4*8 \pmod{13} \\ & b &\equiv& 32 \pmod{13} \\ & b &\equiv& 32-2*13 \pmod{13} \\ & b &\equiv& 6 \pmod{13} \\ & \mathbf{ b } &\equiv& \mathbf{ 6 \pmod{13} } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline 2ab+bc+ca &\equiv& 0\pmod{13} \quad & | \quad b=6, \ c=9 \\ 2a*6+6*9+9*a &\equiv& 0\pmod{13} \\ 21a+54 &\equiv& 0\pmod{13} \quad & | \quad 21\equiv 8\pmod{13},\ \quad 54\equiv 2\pmod{13} \\ 8a +2 &\equiv& 0\pmod{13} \quad & | \quad :2 \\ 4a+1 &\equiv& 0 \pmod{13} \quad & | \quad -1 \\ 4a &\equiv& -1 \pmod{13} \quad | \quad : 4 \\ a &\equiv& (-1)*\dfrac{1}{4} \pmod{13} \\ && \boxed{\dfrac{1}{4} \pmod{13} \\ \equiv 4^{\varphi(13)-1}\pmod{13} \\ \equiv 4^{12-1}\pmod{13} \\ \equiv 4^{11}\pmod{13} \\ \equiv 4194304\pmod{13} \\ \equiv 10\pmod{13} } \\ a &\equiv& (-1)*10 \pmod{13} \\ a &\equiv& -10 \pmod{13} \\ a &\equiv& 13-10 \pmod{13} \\ \mathbf{ a } &\equiv& \mathbf{ 3 \pmod{13} } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && \mathbf{a+b+c \pmod{13}} \\ &\equiv& 3+6+9 \pmod{13} \\ &\equiv& 18 \pmod{13} \\ &\equiv& 18-13 \pmod{13} \\ &\equiv& \mathbf{ 5 \pmod{13} } \\ \hline \end{array}\)