Pure Mathematics - Proof (3)

1. It is known that x and y are integers such that 3x^2 - 5y^2 is divisible by 4.
(a) show that x is even or y is even.
(b) show that both x and y must be even.
(a)
3x^2 - 5y^2 is divisible by 4
if x and y are both odd
let x=2m+1, y=2n+1
then
3x^2 - 5y^2
=3(2m+1)^2-5(2n+1)^2
=3(4m^2+4m+1)-5(4n^2+4n+1)
=12m^2+12m-20n^2-20n-2
but it is not divisible by 4
So x is even or y is even
(b)
if only x is even
let x=2m, y=2n+1
then
3x^2 - 5y^2
=3(4m^2)-5(4n^2+4n+1)
=12m^2-20n^2-20n-5
which is not divisible by 4
if only y is even
let x=2m+1, y=2n
then
3x^2 - 5y^2
=3(4m^2+4m+1)-5(4n^2)
=12m^2+12m-20n^2+3
which is not divisible by 4
So both x and y must be even
2
(a)
diophantine equation x^2 - y^2 = 10
(x-y)(x+y)=10
Since both x and y are positive integer
x+y>x-y
The possible answer are
x+y=10, x-y=1
x+y=5,x-y=2
but they cannot give the positive integer solutions of x,y
So there are no positive integer solutions to the diophantine equation x^2 - y^2 = 10
(b)
if there is a rational solution (p/q) to the equation f(x)=x^5 + x^4 + x^3 + x^2 + 1 = 0
then we have p|1 or -1 , q|1 or -1
Reference
http://purplemath.com/modules/rtnlroot.htm
So the possible rational solution is 1,-1
However
f(1)=5, f(-1)=1.
So 1,-1 are not the roots of f(x) and we conclude that there is no rational solution to the equation x^5 + x^4 + x^3 + x^2 + 1 = 0