Prove that if k and ℓ are posetive integers then k^2 − ℓ^2 can never be equal to 2.

Solution

Proof

If k² − ℓ² has to be 2 then either both k and ℓ has to be even or both have to be odd.

If one is even and the other is odd then k² − ℓ² would be odd.

Now we solve case wise depending on whether both are odd or both are even.

Case 1: Both k and ℓ are even.

Then say k = 2m and ℓ = 2n when m and n are positive integers.

Then k² − ℓ² = (2m)² − (2n)²

= 4m² − 4n²

= 4(m² − n²).

Now since (m² − n²) is an integer and 4 times an integer cannot be 2 so k² − ℓ² cannot be 2 in this case.

Case 2: Both k and ℓ are odd.

Then say k = 2m + 1 and ℓ = 2n + 1 when m and n are positive integers.

Then k² − ℓ² = (2m + 1)−(2n + 1)²

= (4m² + 4m + 1) − (4n² + 4n + 1)

= 4(m² + m − n² - n).

Now since (m² + m − n² - n) is an integer and 4 times an integer cannot be 2 so k² − ℓ² cannot be 2 in this case.