Let $a+2=b=c-2$ with $b$ odd. Then $a^2+b^2+c^2=(b-2)^2+b^2+(b+2)^2=3b^2+8$ And the possible results are: $1111, 3333,5555,7777,9999$ And checking mod $3$ for $4,0,8,4,0$ For the solution $\boxed{41,43,45}$

It's either $1111$ or $5555$ because of $\pmod{10}$. Then u just estimate with $\frac{1111}{3} = 370$, because $(x-2)^2 + x^2 + (x+2)^2 = 3x^2 + 8 \approx 3x^2$ so $17^2 + 19^2 + 21^2$. Not possible, check $5555$ which gives around $43^2$ so $41^2 + 43^2 +45^2$, which works.

Let the numbers be $2n-1$, $2n+1$ and $2n+3$, then $(2n-1)^2+(2n+1)^2+(2n+3)^2=12n^2+12n+11$. If this is a four digit number with four equal digits, then it is a multiple of $11$, so $0\equiv 12n^2+12n+11\equiv n^2+n\equiv n(n+1)\pmod{11}$. Hence $n\equiv 0\pmod{11}$ or $n\equiv -1\pmod{11}$. Now, $n=11$ and $n=22$ do not work, with $n=33$ we get that $12n^2+12n+11$ has more than four digits. We try $n=10$ and then $n=21$ to find $41^2+43^2+45^2=5555$.

Let $a= 2k-1$, $b= 2k+1$ and $c= 2k+3$, with $k\in\mathbb{Z+}$. So, we know that $a^2+b^2+c^2= \overline{xxxx}$, with $x\in$ $\{1,2,3,4,5,6,7,8,9\}$. $\implies (2k-1)^2+(2k+1)^2+(2k+3)^2= 1111x$ $\implies 4k^2-4k+1+4k^2+4k+1+4k^2+12k+9= 1111x\implies 12(k^2+k+1)= 1111x+1$(*) Then, we have two consequences: $\begin{cases} 2\mid 1111x+1\implies 1111x+1 \mbox{ is even}\implies x \mbox{ is odd} \mbox{ (1)} \\ 3\mid 1111x+1\implies 1111x\equiv -1(mod \mbox{ 3})\implies x\equiv -1(mod \mbox{ 3})\mbox{ as } x\in \{1,2,3,4,5,6,7,8,9\}\implies x= 2, 5 \mbox{ or } 8 \mbox{ (2)} \end{cases}$ From (1) and (2), we have that $x= 5$. So, in (*), we will get: $12(k^2+k+1)= 5556\implies k^2+k+1= 463\implies k^2+k-462=0$. Solving, we have that $k_1= \frac{-1- 43}{2}= -22$ and $k_2= \frac{-1+43}{2}= 21$. As $k>0\implies k= 21$. Therefore, the triple $(a,b,c)= (41, 43, 45)$