Is it possible?

From the given conditions we have:

$n^2= 0 \mod a$ (1) and ${(n+1)}^2= 0 \mod a$ (2).

(2) can be simplified to $2n = -1\mod a$ after expanding and using (1).

If $a$ is prime, then (1) only has the solution $n=0 \mod p$ (3). Substituting into (2) leads to $0= -1 \mod p$ , which is not true.

Now consider composite $a$ , with prime factorization $1^{c_0}×{p_1}^{c_1}×{p_2}^{c_2}×...×{p_i}^{c_i}$ . The solutions to (1) must be of the form $1^{c_0}×{p_1}^{c_1}×{p_2}^{c_2}×...{p_b}^{ \frac {c_b} {2}}×...×{p_i}^{c_i}$ , with $0 \leq b \leq i$ and even $c_b$ (as odd $c_b$ would result in non integral $n$ , which is prohibited by the given conditions), leading to $n ={p_1}^{c_1}×{p_2}^{c_2}×...{p_b}^{ \frac {c_b} {2}}×...×{p_i}^{c_i} \mod a$ (4). If $b=0$ , then the result is (3), which does not solve the problem. If $1 \leq b \leq i$ , then after substituting into (2), $2×({p_1}^{c_1}×{p_2}^{c_2}×...{p_b}^{ \frac {c_b} {2}}×...×{p_i}^{c_i})=-1 \mod a$ . Now, all ${p_b}^{\frac {c_b} {2}} \geq 2$ (with the exception of $b=0$ , covered above). This means that $a \geq 2×({p_1}^{c_1}×{p_2}^{c_2}×...{p_b}^{ \frac {c_b} {2}}×...×{p_i}^{c_i})$ , and the difference between them will never be $1$ (try very low values of ${p_b}^{\frac {c_b} {2}}$ ). This means that $2n$ will never be $-1 \mod a$ , and therefore (1) and (2) cannot both hold.

To prove the same result for $n^3$ and ${(n+1)}^3$ , (3) holds, and for composite $a$ , the general form of the solutions is $1^{c_0}×{p_1}^{c_1}×{p_2}^{c_2}×...{p_b}^{ \frac {c_b} {3}}×...×{p_i}^{c_i}$ ., with $0 \leq b \leq i$ and $c_b$ divisible by $3$ .

While $1$ is neither prime nor composite, I included it in my prime factorization in order to make the general form cover all solutions.

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