Let $T_n=\{11(k+h)+10(n^k+n^h)\mid1\leq k,h \leq 10\}$ for $n\in\mathbb N$. Find all $n$ such that there is no $a\neq b\in T_n$ with $110 \mid (a-b)$. With every positive integer $n$, we have a set $T_n=\{11(k+h)+10(n^k+n^h) \mid 1\leq k,h \leq 10\}$. Find all $n$ such that there is no $a\neq b\in T_n$ with $110 \mid (a-b)$.
I tried to divided it into some case with $n\leq 11$, and then to discard the case not right. Is there some other way to solve it?
 A: We can solve this without resorting to brute force.

Firstly, define, for positive integer $n$ and $1\leq k,h\leq10$, $f_n(k,h)=11(k+h)+10(n^k+n^h)$. Then the condition $\exists a,b\in T_n$ such that $a\ne b$ and $a\equiv b\pmod{110}$ is equivalent with the conditions below.
$$\begin{cases}
n^{k_1}+n^{h_1}&\equiv n^{k_2}+n^{h_2}\pmod{11}\\
k_1+h_1&\equiv k_2+h_2\pmod{10}\\
f_n(k_1,h_1)&\ne f_n(k_2,h_2)
\end{cases}.$$
If $(k_1,h_1,k_2,h_2)$ satisfies the above congruences, then call it a valid pair.

If $n^5\equiv1\pmod{11}$, then $n^1\equiv n^6\pmod{11}$ and $n^2\equiv n^7\pmod{11}$. Also $1+2\equiv6+7\pmod{10}$, so such an $n$ does not satisfy the conditions. By the same token, if $n\equiv-1\pmod{11}$ then $n$ does not satisfy the condition. Hence all that rest are those $n$ such that if $n^k\equiv1\pmod{11}$ then $10\mid k$, i.e. primitive roots of $11$, which are $\equiv2,6,7,8\pmod{11}$.
So let $n$ be a primitive root modulo $11$.
Then the condition $n^{k_1}+n^{h_1}\equiv n^{k_2}+n^{h_2}\pmod{11}$ becomes 
$$n^{k_1}+n^{h_1}\equiv n^{k_2}+n^{k_1+h_1-k_2}\pmod{11}.$$
Hence $n^{k_2}$ satisfies the equation $$x^2-(n^{k_1}+n^{h_1})x+n^{k_1+h_1}\equiv0\pmod{11}.$$
But $x^2-(n^{k_1}+n^{h_1})x+n^{k_1+h_1}=(x-n^{k_1})(x-n^{h_1})$, thus $n^{k_2}$ is congruent modulo $11$ to either $n^{k_1}$ or $n^{h_1}$. Since $n$ is a primitive root modulo $11$, this implies that $k_2=k_1$ or $k_2=h_1$. This shows that $f_n(k_1,h_1)=f_n(k_2,h_2)$, and hence $(k_1,h_1,k_2,h_2)$ is not a valid pair.
Therefore indeed for $n\equiv2,6,7,8\pmod{11}$, there are no valid pairs, i.e. such $n$ are what we are looking for.

Hope this helps.
A: We can find $n\pmod{110}$. Here is a simple Python code(maybe it can be optimized, comment about optimization are welcome) you can use to find all $n\pmod{110}$. 
In this program, first storing the values in the set $T_n$ using t(n) function in the program. Then choosing an element $a$, another element $b$ from the rest of the set. But we can have same value again, so checking if $a$ is not equal to $b$, then if $110$ does not divides $(a-b)$, storing that $n$ in an array and printing the array as a set. 
The $n$ values for which $T_n$ satisfies the property are: 
$$n\equiv 2, 6, 7, 8, 13, 17, 18, 19, 24, 28, 29, 30, 35, 39, 40, 41, 46, 50, 51, 52, 57, 61, 62, 63, 68, 72,\\ 73, 74, 79, 83, 84, 85, 90, 94, 95, 96, 101, 105, 106, 107\pmod{110}$$
def t(n):
  arr=[]
  for k in range(1,11,1):
    for h in range(1,11,1):
      val=11*(k+h)+10*(n**k+n**h) 
      arr.append(val) 
  return arr 

st=[]
for n in range(1,110): 
  arr=t(n) 
  flag=0
  for i in range(len(arr)): 
    for j in range(i+1,len(arr)):  
      if(arr[i]!=arr[j]): 
        if((arr[i]-arr[j])%110==0): 
          flag=1
          break
    if(flag==1): 
      break 
  if(flag==1):
    continue
  st.append(n)
print(set(st)) # prints all n in that range
print(len(st))

You can run it online here by pasting the above code and choosing Python 3 from left column there. After pasting press the Run button.

Edit:(thanks to @John for his valuable comment)
Suppose $n_0$ is a solution. Then as in the second term we have a factor of $10$, so, if $n_1\equiv n_0\pmod{11}$ we have $n_1$ also a solution. Hence, finding $n$ below $11$ will be enough. As we can see from above code, $2,6,7,8$ are the only $n$ below $11$, hence any number $n$ below $110$ which is congruent to one of $2,6,7,8\pmod{11}$ is a solution. 
