let x,y be elements of integers. proove that if xy and x+y are even, then both x and y are even.?

Answer 1

both x and y will always be even if xy and x+y are even. if one of the numbers (x, for example) was odd, it would throw things off.

Odd x Odd = Odd
Odd x Even = Even
Even x Even = Even

Odd + Odd = Even
Odd + Even = Odd
Even + Even = Even

If both numbers were odd, xy would be odd.
If there was one odd and one even, x + y would be odd.

But if both numbers are even, both xy and x+y would be even.

Therefore, if xy and x+y are even, both numbers must be even.

Answer 2

Sounds like a vector space / modular arithmetic / linear diophantine equation problem.

xy = even => xy = 0 (mod 2)
x+y = even => x+y = 0 (mod2)

So using the congruence relation of modular arithmetic states....

if a1 cngr b1 (mod n) and a2 congr b2 ( mod n) then

(a1 + b1) congr (b1 + b2)( mod n)

and

(a1 b1) congr (b1 b2) (mod n)

...

In your case, mod n would be mod 2 => 'even'
b1 and b2 would be 0 since division of even numbers by 2 has a remainder of 0

and a1 a2 are x and y in your example...

The source listed below gives an excellent explanation

Answer 3

i make you right!!i dont give a s h it what x or y is!! you can go to the top of your class!! you sad boy!!

Answer 4

x+y even => x,y both even or x,y both odd
xy even => x,y both even or x,y one of each

the only wat to satisfy both of these is for x,y to both be even