<s>[INST] Modular arithmetic - How to find failure case theoretically I am unable to find the wrong intersection point of two modular classes theoretically, for the below question:

If $n$ is an odd integer, and $n^2 + 2n$ and $n$ have the same residue $\pmod 8$. What can be the remainder on division by $8$? 

Residues of $n \mod 8$, for odd $n$, are: $\{1,3,5,7\}$. And of $2n \mod 8$ the residues are: $(2,6,2,6)$, and of $n^2 \mod 8$ are found by simply finding $r^2 \mod 8$:$(1,1,1,1)$, or $\{1\}$. $n^2 + 2n$ have the possible residues found by adding up the corresponding values of residues for $2n \pmod 8$ and $ n^2 \pmod 8$: $(3,7,3,7)$, or the set: $\{3,7\}$. So, $n$ values can be $8k+3, 8k+7, \forall k \in \mathbb {Z}$. So, the remainders $\mod 8$ can be : $3, 7$.
But, there is a flaw in this approach: you cannot figure out that $8k+3$ case will not work before-hand. One need see this case manually, with $n \pmod 8$ yielding residue $3$ and $n^2 + 2n \pmod 8$ yielding residue (using residue rather than actual value of $8k+3$): 
$3^2 + 6 => 15 \pmod 8 => 7$.
This is also substantiated by a value of $n=19$ that falls in the class $8k+ 3$, with the value of $n^2 + 2n = 399 \equiv 7 \pmod 8$. However, $19 \equiv 3\pmod 8$.

Addendum Based on the answers given, have modified the answer to : 
Need to equate the solutions given (theoretically, or algebraically) for the two modular classes:
$n^2 + 2n \equiv n\pmod 8 => n + 2  \equiv 1\pmod 8 => n \equiv -1 \pmod 8$.

Addendum -2 My approach for modification is to theoretically factor in the residue(remainder) from the start, as follows for the OP case, with $k$ representing not $n$ but the residue.
$n^2 \mod 8$ is given for odd $n$ as: $(2k+1)^2 \mod 8 => (4k^2 + 4k + 1) \mod 8 $. On adding $2n \mod 8$ to it, get: $((4k^2 + 4k + 1) + (4k+2)) \mod 8 => (4k^2 + 8k + 3) \mod 8$. This should be equal to $n \pmod 8$. 
$4k^2 + 8k + 3 \equiv 2k+1 \pmod 8 => 2k^2 + 3k + 1 \equiv 0 \pmod 8 => 2k(k+1) + 1.(k+1) \equiv 0 \pmod 8 => (2k+1)(k+1) \equiv 0 \pmod 8$.
Only answer for integer value of $k$ (residue) possible is: $-1$ in modulo $8$ system
 [/INST] All odd numbers square to $1$ mod $8$ so your equation actually reads $$1 + 2n \equiv n \, \text{mod} \, 8,$$ with solution $n \equiv 7$.
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