Question

if 4 divides a^2-3b^2, then at least one of the integers a and b is even.

Answer 1

Answer:

Step-by-step explanation:

Suppose, on the contrary, a and b are both odd integers, that is:

$a=2n+1\\\\b=2m+1$

m and n being some integers numbers.

This way you have to:

$a^{2}-3b^{2}=(2n+1)^2-3(2m+1)^2=(4n^2+4n+1)-3(4m^2+4m+1)\\\\a^{2}-3b^{2}=4(n^2-3m^2+n-3m)-2=4(n(n+1)-3m(m+1))-2$

The last expression cannot be divisible by 4 since 2 is not divisible by 4. The previous conclusion leads to a contradiction, which was generated from the assumption that a and b were both odd integers. In conclusion, at least one of the two a and b should be an even integer