Problem
NT-B1-M03-P023 Two Squares Divisible by 7
#23
★★★★☆ Level 4 of 5
Prove: if \(7\mid x^2+y^2\), then \(7\mid x\) and \(7\mid y\).
Squares modulo \(7\) are \(0,1,2,4\).
The quadratic residues modulo \(7\) are \(0,1,2,4\). For the sum of two such residues to be \(0\), the only possibility is \(0+0\): the negatives of \(1,2,4\) are \(6,5,3\), which are not square residues. Thus \(x^2\equiv0\) and \(y^2\equiv0\pmod7\), so \(7\mid x\) and \(7\mid y\).
A good example where one must inspect not only the table, but sums of residues.