(In reply to

Generalised approach by Harry)

Excellent finding, Harry!

I thought that your main points got buried, so I am reposting them. (No original work on my part was done)

You astutely wrote:

So for a_{24} to be an integer, (a_{2} – a_{1}) must be divisible by 2^{22}

If only positive integers are allowed for (a_{1}, a_{2}) then choose (1, 2^{22} + 1).

If zero is allowed then (0, 2^{22}) is best, as Jer recommended.

If all integers are allowed then (-2^{21}, 2^{21}) have the smallest abs. values