Question

Suppose $x-3$ and $y+3$ are multiples of $7$. What is the smallest positive integer, $n,$ for which $x^2+xy+y^2+n$ is a multiple of $7$? Enter your answer. I need Immediate help or you wont get the points.

Answer 1

In the language of modular arithmetic, we're given

`x-3\equiv0\pmod7\implies x\equiv3\pmod7`

`y+3\equiv0\pmod7\implies y\equiv-3\equiv4\pmod7`

Then x = 7a + 3 and y = 7b + 4 for integers a and b.

Substitute these into the quadratic expression and simplify:

`x^2+xy+y^2+n\equiv0\pmod7`

`(7a+3)^2+(7a+3)(7b+4)+(7b+4)^2+n\equiv0\pmod7`

`49a^2+42a+9+49ab+28a+21b+12+49b^2+56b+16+n \equiv 0\pmod7`

`37+n\equiv 0\pmod7`

`n\equiv-2\equiv5\pmod7`

which means the smallest positive integer n we are looking for is 5.