.suppose a and b are integers and a^2-5b is even. Prove that b^2-5a is even.

Answer 1

a and b are integers.

and a² - 5b is even.

Consider the cases.

1. a and b are even.

2. a and b are odd.

3. a is even and b is odd.

4. a is odd and b is even.

Case(1) a and b are even then

 a² - 5b

= (even)² - 5(even)

= even -even

= even.

For Example

a = 4 and b = 2 then

= (4)² - 5(2)

= 16 - 10

= 6

is a even integer.

Case(2) a and b are odd then

 a² - 5b

= (odd)² - 5(odd)

= odd -odd

= even.

For Example

a = 3 and b = 1 then

= (3)² - 5(1)

= 9 - 5

= 4

is a even integer.

Case(3) and Case(4) are not considered as it does not satisfies the condition a² - 5b is even.

Now we prove that b² - 5a is even when a² - 5b is even.

Case(1) a and b are even then

 b² - 5a

= (even)² - 5(even)

= even -even

= even.

For Example

a = 4 and b = 2 then

= (2)² - 5(4)

= 4- 20

= -6

is a even integer.

Case(2) a and b are odd then

 b² - 5a

= (odd)² - 5(odd)

= odd -odd

= even.

For Example

a = 3 and b = 1 then

= (1)² - 5(3)

= 1 - 15

= -14

is a even integer.

Therefore it is proved that  b² - 5a is even when a² - 5b is even.