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

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.