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


closed as off-topic by Claude Leibovici, Shaun, 3SAT, N. F. Taussig, Daniel W. Farlow Feb 13 '16 at 13:50

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Claude Leibovici, Shaun, 3SAT, N. F. Taussig, Daniel W. Farlow
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ yes , sorry for my typo $\endgroup$ – zeraoulia rafik Feb 13 '16 at 7:58

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.

  • $\begingroup$ thank you very much for this answer $\endgroup$ – zeraoulia rafik Feb 13 '16 at 8:22

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.


Not the answer you're looking for? Browse other questions tagged or ask your own question.