Yearly sequences

Note that the sequence has exactly $2015$ terms and each term is a non-zero integer. So, the square of each term of the sequence must be a positive integer.

There are $2015$ terms in the sum and each term contributes positively to the sum. If we denote the general term of the sequence as $a_i$ , we make the following claim:

Claim: For the conditions given in the problem to hold, we must have $a_i=1,(-1)$

Proof: We already know that $(a_i)^2\gt 0$ is a positive integer. So, the minimum value it can take is $1$ . The minimum of the given sum is attained when the square of each of the terms of the sequence contribute the minimum possible value to the sum. And that minimum sum is $2015$ when each of the term contribute the minimum positive value of their square which is $1$ . So, we have,

$\sum_{i=1}^{2015} \left((a_i)^2\right)\geq 2015\times 1=2015$

and equality holds iff $a_i=1,(-1)~\forall~1\leq i\leq 2015$

Now, assume that any one of the values or multiple values of the sequence isn't $1,(-1)$ . Then that must mean that the square of those term(s) is(are) greater than $1$ , i.e., the required sum is also greater than $2015$ .

Hence, the claim is proved.

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Since we want the equality case, we have only $2$ choices for $a_i$ and repetition is allowed, so by the rule of product, we have,

$\boxed{\textrm{No. of such sequences }=2^{2015}}$

Troll question. Isn't worth the Level 4 Rating it has got. :p

$|a_i| \geq 1$

$\implies a_i^2 \geq 1$

$\implies \sum a_i^2 \geq 1 + 1 + 1 + ... + 1 + 1$ ,

$\implies \sum a_i^2 \geq 2015$ .

Since $\sum a_i = 2015$ , this means that $a_i^2 = 1$ for all $i$ . So for all $i$ , $|a_i| = 1$ or $a_i = 1,-1$ .

For each $i$ , $a_i$ can be $1$ or $-1$ . Since there are 2015 terms in the sequence, by rule of product, the total number of sequences is $2^{2015}$ and $a+b = \boxed{2017}$