Give a counterexample showing that \(\left(\frac{a+b}{p}\right)=\left(\frac{a}{p}\right)+\left(\frac{b}{p}\right)\) fails in general, taking \(p=7\), \(a=1\), \(b=2\).
Choose one option.
Show answer
C. It fails: \(\left(\frac{3}{7}\right)=-1\) while the sum gives \(1+1=2\)
Since \(\left(\frac{1}{7}\right)=1\) and \(\left(\frac{2}{7}\right)=1\), their sum is 2, but \(\left(\frac{3}{7}\right)=-1\), because \(3\) is not among the nonzero squares \(1,2,4\); the equality fails.
The Legendre symbol \(\left(\frac{a}{p}\right)\) for an odd prime \(p\) and an integer \(a\) with \(p \nmid a\) equals \(1\) if \(a\) is a quadratic residue modulo \(p\), and \(-1\) if it is not.