2011/009) one problem in AP

question

The fourth power of the common difference of an AP with integer entries is added to the product of any four consecutive terms of it. Prove that the resulting sum is the square of an integer.

proof

With out loss of generality we can take the 1st term of the 4 consecutive terms to be a and let the difference be t

a and t are integers

We have the 4 terms = a, a+t, a+2t, a+ 3t and 4th power of difference is t^4

Now a(a+t)(a+2t)(a+3t) + t^4

= (a(a+3t))((a+t)(a+2t)) + t^4
= (a^2+3ta)(a^3+3ta + 2t^2) + t^4

Letting a^2 + 3ta = p we get

= p(p+2 t^2) + t^ 4= (p^2+ 2pt^2+t^4) = (p+t^2)^2 = (a^2+3at+t^2)^2

Which is square of integer as a^2+3at+t^2 is integer