Find a second-degree polynomial with integer coefficients, p(x) = ax² + bx + c, such that p(1), p(2), p(3), and p(4) are perfect squares (that is, squares of integers), but p(5) is not.

I did a search based on varying the values for p(1),p(2), and p(3) amount the first ten perfect squares.  For each of these combinations, I solved for the interpolating polynomial as these 3 are guaranteed to uniquely determine a second-degree polynomial.  I then tested to first make sure that a is not zero, a b and c are all integers, and that f(4) is square while f(5) is not.  These gave a lot of solutions most of which had a pattern where p(1)=p(4) and p(2)=p(3), so I decided to try and find one that is a bit more interesting where all of p(1), p(2), p(3), and p(4) are unique as well as having p(x) having no factorization over the rationals.  A quick search of my results gave this solution:
p(x)=57x^2-210x+217 which gives
p(1)=64=8^2
p(2)=25=5^2
p(3)=100=10^2
p(4)=289=17^2
p(5)=592=37*4^2 and thus is not a square number

Posted by Daniel on 2013-08-09 22:38:37