(a): By performing $n \to n-2015$, and changing $b,c,d$ appropriately, we can just go ahead and look at $a_0, a_1$. First, we give a construction for (a). Take $\boxed{a_n=n(n^2-n+1)}$. Clearly $a_0=0$ and $a_1=1$. As $n^2-n+1 > 0$, $a_n$ cannot be a square for $n<0$. Also, as $(n,n^2-n+1)=1$, both $n$ and $n^2-n+1$ must be a square. However, since $(n-1)^2 < n^2-n+1 < n^2$, $n^2-n+1$ cannot be a square for $n \ge 2$, so this gives a construction.

. Let $a_0=u^2$ and $a_1=v^2$. Now look at points $(0,u)$ and $(1,v)$. The line between these two is $y=(v-u)x+u$. Let's calculate the intersection between $y^2=x^3+bx^2+cx+d$ and $y=(v-u)x+u$. We use Vieta's relations. Solving for $x$ gives $$x^3+(b-(v-u)^2)x^2+(c-2u(v-u))x+(d-u^2)=0$$so Vieta's relations give that the third point is $$(-b+(v-u)^2-1, (v-u)((v-u)^2-b-1)+u)$$which clearly has integer coordinates. This coordinate must be equal to one of $(0,u)$ or $(1,v)$. This gives us $(v-u)^2=b+1$ or $(v-u)^2=b+2$. As the graph of $y^2=x^3+bx^2+cx+d$ is symmetric wrt the $x$-axis, we can look at $(0,-u)$ and $(1,v)$. This will gives us $(v+u)^2=b+1$ or $(v+u)^2=b+2$. By parity, we have $(v+u)^2=(v-u)^2$, or $4uv=0$, or $uv=0$, or $a_0a_1=0$. This gives us that the only possible value for $a_{2015} \cdot a_{2016}$ is $\boxed{0}$.

Of course, we replace $a_{2015}$ and $a_{2016}$ with $a_0$ and $a_1$ everywhere. (Like, really?) For part (a), one suitable example is $a(n) = n(n^2+7n+1)$, since the two factors are relatively prime and $n^2+7n+1$ is never zero, and a square only when $n=1$. Hence $a_0 = 0$ and $a_1 = 9$. For part (b) we claim the answer is $a_0a_1 = 0$ (construction done above). Assume $a_0 = v^2$ and $a_1 = (u+v)^2$, and consider the polynomial \[ n^3+bn^2+cn+d - (un+v)^2. \]It has a root at $n \in \{0,1\}$ and if it has any other root we get another $n$ with $a(n)$ a perfect square. Consequently, if $a(n)$ has the desired property we must have \[ n^3+bn^2+cn+d-(un+v)^2 \equiv n^2(n-1) \text{ or } n(n-1)^2. \]Thus \[ a(n) = n^2(n-1) + (un+v)^2 \quad\text{ or }\quad a(n) = n(n-1)^2 + (un+v)^2. \]But now in the former case plugging in $n = 4v(u+v)$ gives yet another square solution; in the latter case $n = 1+4v(u+v)$. Thus we must have $0 \in \{v, u+v\}$ as desired.

Replace $a_{2015}$ and $a_{2016}$ with $a_0$ and $a_1$. For part $a$, just take $a_n = n(n^2-n+1)$. The two factors are relatively prime, so if they multiply to a square, $n^2-n+1$ must be a square, contradiction for $n>1$. For part $b$, we use elliptic curves. Consider the curve $y^2 = x^3 + ax^2 + bx + c^2$ with $c>0$. Suppose $a_1 = d^2$ with $d>0$. Consider the two polynomials $$x^3 + ax^2 + bx + c^2 - ((d-c)x + c)^2 = 0 \quad (1)$$$$x^3 + ax^2 + bx + c^2 - ((d+c)x - c)^2 = 0\quad (2)$$Note that this basically finds the intersection points of the lines thru $(0, \pm c)$ and $(1,d)$ with the curve $y^2 = x^3 + ax^2 + bx + c^2$. For equation $(1)$, the roots are $0, 1,$ and $(d-c)^2 - a$ by Vieta. If $(d-c)^2 - a\neq 0,1$ then note that $a_{(d-c)^2 - a}$ is the square of an integer - in particular, if $t = (d-c)^2 - a$ then $a_t = t^3 + at^2 + bt + c^2 = ((d-c)t + c)^2$. Thus $(d-c)^2 - a\in \{0,1\}$. But similarly $(d+c)^2 - a\in \{0,1\}$. Thus $|4cd|\le 1$. Since $c=|a_0|$ and $d=|a_1|$ are integers, it follows that $|a_0a_1|=0$, as desired. $\square$

Note that by shifting we have the following mapping: $a_{2015}\mapsto a_{0}$, $a_{2016}\mapsto a_1$. Consider $y=(rx+s)^2$. Put $a_0=s^2$ and $a_1=(r+s)^2$; it follows that the polynomial $$y=x^3+bx^2+cx+s^2-(rx+s)^2=x^3+(b-r^2)x^2+(c-2rs)x$$with $1+b+c=r^2+2rs$ has a double root. We can transform our polynomial to $$y=x^3+(b-r^2)x^2+(r^2-b-1)x$$to obtain our $x$ value roots as $0,1,r^2-b-1$. The abscissa $x=r^2-b-1$ either corresponds to $r^2=b+1, r^2=b+2$. This means our difference polynomials are simply $y=x^3-x^2$, $y=x^3-2x^2+x$, so our overall polynomials are $$y=(rx+s)^2+x^3-x^2, y=(rx+s)^2+x^3-2x^2+x$$Selecting $x=4s(r+s)$ in the first and $x=4s(r+s)+1$ yields $y=s^2(4(r+s)(r+2s)-1)^2$ in the former and $y=(4r^2s+12rs^2+r+8s^3+s)^2$ in the latter, which forces one of $s,r+s=0$. Hence, the product is always $\boxed{0.}$ Finally, we claim that $a_n=n^3-n^2+n$ is a valid cubic sequence. Indeed, note that $a_n$ factors as $n(n^2-n+1)$, and since these two factors are coprime, it follows that both must attain perfect square values. It is easy to see by bounding that $n^2-n+1$ only does so for $n=0,1$, whence the conclusion follows$.\:\blacksquare\:$

The main idea is to observe that it almost never happens that only two integer values of $n$ yield a perfect square, as we can intersect the line through these two points again with the cubic to get a third integer solution. Then it is simply a matter of determining when the third intersection coincides with the initial two points. This idea is very much motivated through the theory of elliptic curves - perhaps this is how the problem was created?

leminscate wrote: The main idea is to observe that it almost never happens that only two integer values of $n$ yield a perfect square, as we can intersect the line through these two points again with the cubic to get a third integer solution. Then it is simply a matter of determining when the third intersection coincides with the initial two points. This idea is very much motivated through the theory of elliptic curves - perhaps this is how the problem was created? I agree, since square=cubic is very much motivated by the theory behind elliptic curves (even though the cubic in question is not necessarily elliptic).

........

rkm0959 wrote: (a): A stupied question:if you turn a(n) into b(n-2015),than how are a(1),a(2),...,a(2014)?

By shifting, it is equivalent to solve the problem with $a_{2015}$ and $a_{2016}$ replaced by $a_0$ and $a_1$ respectively. For part (a), take $a_n=n(n^2-n+1)$, so $a_0=0$ and $a_1=1$. Since $\gcd(n,n^2-n+1)=1$, if this expression is a square then both terms must be squares as well. On the other hand, for $n\geq 2$ we have $(n-1)^2<n^2-n+1<n^2$, and for $n \leq -1$ we have $n^2<n^2-n+1<(n+1)^2$, so there are no other square terms in the sequence. For part (b), the only answer is $0$, which can be achieved with the construction already given. Suppose that the monic cubic polynomial $p(x)$ takes on squares for $x=0,1$, but not other $x$. Let $p(0)=b^2$ and $p(1)=a^2$, where $a,b \neq 0$. Then consider the polynomial $$p(x)-((a-b)x+b)^2,$$which is $0$ for $x=0,1$. This means that its third root $r$ must also be an integer (by polynomial division). But if this root is not $0$ or $1$, then $p(r)$ is a square as well, specifically $((a-b)r+b)^2$: contradiction. Therefore $r \in \{0,1\}$ and we must actually have $p(x)=x^2(x-1)+((a-b)x+b)^2$ or $p(x)=x(x-1)^2+((a-b)x+b)^2$. If $p(x)=x^2(x-1)+((a-b)x+b)^2$, then we can check that $p(4ab)=(4ab(a+b)-b)^2$. Since $4ab \not \in \{0,1\}$, we have a contradiction. If $p(x)=x(x-1)^2+((a-b)x+b)^2$, then we can check that $p(4ab+1)=(4ab(a+b)+a)^2$. Since $4ab+1 \not \in \{0,1\}$, we also have a contradiction. Therefore we find that $a_0a_1$ must be $0$, so we're done. $\blacksquare$ Remark (finding the counterexamples): After essentially determining the form of $p$, I initially thought the answer was everything (after thinking correctly that the answer was $0$ for a few minutes), using $x^2(x-1)+(kx+1)^2$ as a construction, but I was unable to prove it. Then I realized that if I could find two divisors of $x^2(x-1)$ with positive difference equal to $2kx+2$, I could get $p(x)$ to be a square. The main difficulty, in some sense, is to get the difference to be $2$ modulo $x$, but if you actually take $x^2/2$ and $2(x-1)$ then this is taken care of and you're just solving a linear equation in $x$ (interestingly, since the two divisors have to be the same parity, you probably want them to both be even). Then you do this for $x(x-1)^2$ as well, and then generalize for $2((a-b)x+b)$ to produce the above solution. Remark (on subtracting $((a-b)x+b)^2$): The subtraction step is very important. To some degree it was motivated by the idea that if you have two points on an elliptic curve you can kind of use them to find a third, and this problem is similar (I barely know anything about elliptic curves, so this might be totally off). Another idea is that I should not be allowed to make $a_0$ and $a_1$ both $0$ "in general", since I realized that I should be able to divide out and find a third root. So this root must actually be $0$ or $1$ and you heavily restrict $p$ once you take the natural generalization to interpolating a linear polynomial between $\sqrt{a_0}$ and $\sqrt{a_1}$.

Isn't it too easy for RMM 3? In my opinion, it is even easier, than IMO 3, let alone RMM.

Part a): Yes, consider $a_n = n(n^2 - n + 1)$. Notice that $a_0$ and $a_1$ are perfect square. If anything else in the sequence was a perfect square, then we clearly must have $n \ge 0$ as $n^2 - n + 1 > (n-1/2)^2 \ge 0$. Now since $n,n^2 - n+ 1$ are coprime, they must both be perfect squares. Hence $4n^2 - 4n + 4$ and $4n^2 - 4n + 1$ are perfect square. Now if integer squares $a^2$ and $b^2 $ have a difference of $3$, then they must be $1$ and $4$ in some order, implying that $4n^2 - 4n + 1 = 1$, so $n^2 - n = 0\implies n\in \{0,1\}$. Part b): We prove the answer is $0$. Clearly works cuz above. Suppose otherwise and there existed $a_n$ with $a_0 a_1 \ne 0$ satisfying the condition in part a). Let $x = \sqrt{a_0}, y = \sqrt{b_0}$. Now let $P(n) = a_n - ((x+y)n - x)^2 $ and $Q(n) = a_n - ((y-x)n + x)^2$. Clearly $0$ and $1$ are roots of $P$ and $Q$. Now if $n$ is a root of $P$ or $Q$, then $a_n$ is clearly a perfect square. Hence the only roots of $P,Q$ can be $0,1$. Thus, the sum of roots of $P,Q $ are both in $\{1,2\}$. By Vieta's, the sum of roots of $P$ is $(x+y)^2 - b$ and the sum of roots of $Q$ is $(y-x)^2 - b$. Since $(x+y)^2 - b$ and $(y-x)^2 - b$ are the same parity and are in $\{1,2\}$, they must be equal, implying $xy = 0$, contradiction.