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The hard part of this problem is the analysis to limit the number of integers we need to consider:

The procedure followed, starting with integer

nis:Now consider:

writing

p = 4n^2 + 2n + 1:so:

also:

So for

n < -1orn > 3we have:i.e.

16 f(n)lies between 2 consecutive squares, so it cannot be a perfect square.Now suppose

f(n) = q^2for some integerq, then:But

16 f(n)cannot be a perfect square, hence neither canf(n).So we only need to explore values for n = -1, 0, 1, 2, 3, which we can do by hand or by program:

Solution:The number first thought of was 3.