How do i find three successive natural odd numbers for which the sum of their squares can be written in decimal system as :$\overline{xxxx} $? let $a, b,c$ be a successive   natural odd numbers, I would like to know how do i find three successive natural odd  numbers for which  the sum of their squares  : $a²+b²+c²$ can be written in decimal system as :$\overline{xxxx} $ ?
Note : I don't understand what it does meant :The  number can be written in decimal system as $\overline{xxxx} $ ?
Source of question:competition for teachers degree   exercice 1
Thank you for any help 
 A: The only meaning I can attach to $\overline{xxxx}$ is that you expect a result made of 4 equal digits (I can't read your source since it seems in Arabic or something similar).
If $n$ is our middle odd number, the sum of the square of 3 consecutive odd numbers is 
$$
(n-2)^2 +n^2+(n+2)^2 = 3n^2+8
$$
and we want this to be equal to $k\cdot 1111$, where $1\le k\le 9$.
We can actually try the 9 possible values for $k$ and solve for $n$, but we can restrict the possibilities.
Since $n$ is odd, $3n^2+8$ is odd. Thus $k$ must be odd, otherwise $k\cdot 1111$ will be even.
Then we can reduce the equation $3n^2+8 = k\cdot 1111$ modulo 3 and we obtain
$2\equiv k \pmod 3$, since $3n^2$ is obviously 0 mod 3 and $1111\equiv 1\pmod 3$ and $8\equiv 2\pmod 3$.  
Thus $k$ can only be equal to 2, 5 or 8. But we have determined that $k$ must be odd so it can only be $k=5$.
Now we check $3n^2+8 = 5555$ and we find that indeed $n=43$ is the solution, and that $41^2 + 43^2+ 45^2=5555$ as requested.
A: My guess is that it might be looking for a result which is four repeated digits such as $1111$ or $2222$. 
One approach would be to look for an example. Since $57^2+59^2+61^2 \gt 10000$, there are not too many to check.
You could cut the checking down by noting that 


*

*the sum of three odd numbers is odd 

*$(n-2)^2+n^2+(n+2)^2 = 3n^2+8$ 

*only one of $1111,3333,5555,7777,9999$ is $8$ more than a multiple of $3$


You find $5555$ and $\sqrt{\dfrac{5555-8}{3}} = 43$ so you want $41^2+43^2+45^2=5555$ as the solution. 
