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

## Solution

### Proof

If k^{2} − ℓ^{2} 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 k^{2} − ℓ^{2} 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 k^{2} − ℓ^{2} = (2m)^{2} − (2n)^{2}

= 4m^{2} − 4n^{2}

= 4(m^{2} − n^{2}).

Now since (m^{2} − n^{2}) is an integer and 4 times an integer cannot be 2 so k^{2} − ℓ^{2} 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 k^{2} − ℓ^{2} = (2m + 1)−(2n + 1)^{2}

= (4m^{2} + 4m + 1) − (4n^{2} + 4n + 1)

= 4(m^{2} + m − n^{2} - n).

Now since (m^{2} + m − n^{2} - n) is an integer and 4 times an integer cannot be 2 so k^{2} − ℓ^{2} cannot be 2 in this case.