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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

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

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

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

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