Find all triples $(a,b,c)$ such that $a,b,c$ are $3$ consecutive odd positive integers and $a^2+b^2+c^2$ is four-digit number 
Find all triples $(a,b,c)$ such that $a,b,c$ are $3$ consecutive odd positive integers and $a^2+b^2+c^2$ is four-digit number consisting of the same four digits. 

I know that $a=2n+1$ then $b=2n+3$ and $c=2n+5$. Also $(2n+1)^2+(2n+3)^2+(2n+5)^2=zzzz$ which the set of all possibilities would be $1111,2222,3333,4444,5555,6666,7777,8888,9999$ I could also simplify this a bit as $12(n^2+3n)+35=zzzz$ 
Now I'm a little stuck on how to proceed with this problem  
 A: Work mod 4 (always a good thing with squares, since squares are $0$ or $1$ mod $4$).
Since $a, b, c$ are odd, the sum of the squares is odd. Moreover, the sum of the squares is $3$ mod $4$ (since all the squares are individually $1$ mod $4$). That means you've only got 1111, 5555, 9999 to play with.
OK, maybe mod 8 will do better. Squares of odd numbers mod 8 are only 1, so the sum of the squares mod 8 must be 3. That leaves only 5555 as an option.
Now just try it. Trial and error gives you $n=20$ making $5555$. (Or you could just solve $12(n^2+3n)+35 = 5555$.)
A: You can write $12n^2+36n+35 =1111t$ so $12|1111t+1$ so $12|89t-1$ so $12|5t-1$. So obviously t is odd and not dvisible by 3. Thus $t\in \{1,5,7\}$ try each of them and you get an answer.  
A: Even when $z=9$, you have 
$$
9999>3(2n+1)^2, 
$$
so
$$
n<\frac{\sqrt{3333}-1}2=28.3\ldots
$$
Also,
$$
1111<3(2n+5)^2,
$$
which implies
$$
n>\frac{\sqrt{1111}-5}2=14.16\ldots
$$
So you only need to try with $$15\leq n\leq27.$$ The only one satisfying the equality is $n=20$. 
