Answer:
See explanation below
Step-by-step explanation:
First, let's see under which conditions a²- 5b is even.
Case 1: a is even and b is even
If a is even then there exists a k >1 such that a = 2k
If b is even then there exists a j > 1 such that b = 2j
⇒a²- 5b = (2k)² - 5(2j) = 4k²-10j = 2( 2k² - 5j)
Therefore is a and b are even, then a²- 5b is even.
Case 2: a is even an b is odd
If a is even then there exists a k ≥ 1 such that a = 2k
If b is odd then there exists a j ≥ such that b = 2j - 1
⇒a²- 5b = (2k)² - 5(2j - 1) = 4k²- 10j + 5 = 4k²- 10j + 4 + 1 = 2 ( 2k² - 5j + 2) + 1
Therefore if a is even and b is odd a² - 5b is odd.
Case 3 : a is odd and b is odd
If a is odd then there exists a k ≥ 1 such that a = 2k - 1
If b is odd then there exists a j ≥ such that b = 2j - 1
⇒a² - 5b = (2k - 1)² -5 (2j - 1) = 4k²- 4k +1 -10j + 5 = 4k² - 4k -10j + 6 = 2 (2k² -2k -5j +3)
Therefore if a is odd and b is odd, a² - 5b is even.
Case 4: a is odd and b is even
If a is odd then there exists a k ≥ 1 such that a = 2k - 1
If b is even then there exists a j ≥ such that b = 2j
⇒ a² - 5b = (2k -1)² - 5 (2j) = 4k² - 4k + 1 - 10j = 2( 2k²- 2k - 5j ) + 1
Therefore is a is odd and b is even, a² -5b is odd.
So now we know that if a and b are integers and a² - 5b is even, then both a and b are odd or both are even.
Now we're going to prove that b² - 5a is even for these both cases.
Case 1: If a² - 5b is even and a, b are even ⇒ b² - 5a is even
If a is even, then there exists a k≥1 such that a = 2k
If b is even, then there exists a j≥1 such that b = 2j
b² - 5a = (2j)² - 5(2k) = 4j² - 10k = 2 (2j² - 5k)
Therefore, b² - 5a is even
Case 2, If a² - 5b is even and a, b are odd ⇒ b² - 5a is even
If a is odd then there exists a k ≥ 1 such that a = 2k - 1
If b is odd then there exists a j ≥ such that b = 2j - 1
b² - 5a = (2j - 1)² - 5(2k - 1) = 4j² - 2j + 1 -10k + 5 = 4j² -2j -10k + 6 = 2 (2j² - j - 5k +3)
Therefore, b² - 5a is even.
Since we proved the only both cases possible, therefore we can conclude that if a and b are integers and a² - 5b is even, then b² - 5a is even.
