To solve the problem, we proceed as follows:

Step 1: Find (a_1)

For (n=1), the given equation is:
(2S_1 = a_1 + 1^2 + 1)
Since (S_1 = a_1):
(2a_1 = a_1 + 2)
(a_1 = 2)

Step 2: Derive the general term (a_n)

For (n \geq 2):
(2S_n = an + n^2 + 1) ...(1)
(2S{n-1} = a_{n-1} + (n-1)^2 + 1) ...(2)

Subtract (2) from (1):
(2(Sn - S{n-1}) = an - a{n-1} + n^2 - (n-1)^2)
Since (Sn - S{n-1} = a_n):
(2a_n = an - a{n-1} + 2n - 1)
Rearranged: (an = -a{n-1} + 2n - 1)

Let (b_n = a_n - n), then:
(bn = -(a{n-1} - (n-1)) = -b_{n-1})

(b_1 = a_1 - 1 = 1), so (b_n = (-1)^{n-1})

Thus: (a_n = n + (-1)^{n-1})

Answers: (a_1 = 2); (a_n = n + (-1)^{n-1})

(\boxed{2})
(\boxed{n + (-1)^{n-1}})

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