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If $n$ is a positive integer greater than 1 such that $3n+1$ is perfect square, then show that $n+1$ is the sum of three perfect squares.

My work:
$3n+1=x^2$
$3n+3=x^2+2$
$3(n+1)=x^2+2$
$(n+1)=\dfrac{x^2+2}{3}$

I have no clue what to do next. Please help!

Hawk
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  • where did you got his question? Some Olympiad? –  Jan 10 '14 at 12:39
  • no, not olympiad...i got it somewhere...don't remember where.had been trying since then without any luck! – Hawk Jan 10 '14 at 12:42
  • I would suggest you to work out for $2n+1$ case first and then look for $3n+1$... i.e., $2n+1$ is a perfect square then $n+1$ is sum of two perfect squares.... let me tell you this is not so easy and this would take some time.... do not get frustrated... Good luck! –  Jan 10 '14 at 12:45
  • No, the cause for my frustration is, this problem was classified as 'easy'. – Hawk Jan 10 '14 at 12:48
  • I am not aware of your level but this is a "Putnam practice level problem"... Good luck any how! –  Jan 10 '14 at 12:49
  • Thank you!...But, I am not aware of my level either! – Hawk Jan 10 '14 at 12:53

3 Answers3

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We are given that $3n+1 = a^2 $.

We want to show that $n+1$ is the sum of 3 perfect squares.

Note that $a$ is not a multiple of 3.

If $ a \equiv 1 \pmod{3}$, then observe that $9n+9 = 3a^2 + 6 = (a-1)^2 + (a-1)^2 + (a+2)^2$, and hence

$$ n+1 = \left( \frac{a-1}{3} \right)^2 + \left( \frac{a-1}{3} \right)^2 + \left( \frac{a+2}{3} \right)^2. $$

If $ a \equiv 2 \pmod{3}$, then observe that $9n+9 = 3a^2 + 6 = (a+1)^2 + (a+1)^2 + (a-2)^2$, and hence

$$ n+1 = \left( \frac{a+1}{3} \right)^2 + \left( \frac{a+1}{3} \right)^2 + \left( \frac{a-2}{3} \right)^2 $$

Thus the result is true.

The motivation behind the solution is: We want to show that $n+1$ is the sum of 3 squares, and the only thing that we have to work with is $a^2$, and possibly things around it. Since $3a^2$ (the naive sum of 3 squares) is so close to $9n+9$, this suggests that we have some sort of wriggle room. Remembering that we have to account for the factor of 3 then greatly restricts our options.

As an extension, show that $n+3$ can also be written as the sum of 3 perfect squares, using a similar approach.

J. W. Tanner
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Calvin Lin
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  • really neat...and great approach...thank you! can you suggest me some materials or books which will help me with these approaches for practice? – Hawk Jan 11 '14 at 06:28
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If $\displaystyle 3n+1=a^2, (a,3)=1\implies a$ can be written as $\displaystyle3b\pm1$ where $b$ is an integer

So we have $\displaystyle 3n+1=(3b\pm1)^2\implies n=3b^2\pm2b$

$\displaystyle n+1=3b^2\pm2b+1=b^2+b^2+b^2\pm2b+1=b^2+b^2+(b\pm1)^2$

lab bhattacharjee
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This will give the answer : http://www.proofwiki.org/wiki/Integer_as_Sum_of_Three_Squares

You need to prove that $n+1=\frac{x^2+2}{3}$ is not of the form $4^k(8m+7)$ for $k,m\in \mathbb{N}$ and its not that difficult to show this.

For Square modulo 8 we know that $x^2 =0,1,4 \mod{8}$, so $x^2 +2 =2,3,6 \mod{8}$. Since $x^2 +2 =0 \mod{3}$ we obtain that $\frac{x^2 +2}{3} = 1,2,A \mod{8}$ where $A=\frac{2+8a}{3}$, $a$ positive integer. And $A=\psi \mod{8}$ namely, $2=3\psi \mod{8}$ with $\psi\in\{0,1,...,7\}$. The only possible value is $\psi=6$ hence $\frac{x^2 +2}{3} = 1,2,6 \mod{8}$.

If it was of that particular form then:

For $m=0$,$k=0$ it is equal to 7 modulo 8

For $m=0$,$k=1$ it is equal to 4 modulo 8

For $m=0$,$k=2$ it is equal to 0 modulo 8

For $m>0,k>2$ it is equal to 0 modulo 8 So you obtain a contradiction.

Kal S.
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  • are you sure it is not difficult... I do not think so... :O –  Jan 10 '14 at 12:29
  • @PraphullaKoushik.Agreed!I cannot do it easily still now and I am trying! – Hawk Jan 10 '14 at 12:38
  • Please show us your technique, you think it is easy. It will be easy for you to show! – Hawk Jan 10 '14 at 12:39
  • I know it is not easy! :D –  Jan 10 '14 at 12:39
  • Hope my answer is more helpful now. Also I tried to help and my first thought was that is must be easy to show that last part. I might be wrong. So I don't understand the frustration. – Kal S. Jan 10 '14 at 12:55
  • are you sure? because a number of the form $4^n(8m+7)$, put $n=1,m=0$,here the remainder is 4 not 7 – Hawk Jan 10 '14 at 12:55
  • Yes but you get again a contradiction. I edited my answer to include this. – Kal S. Jan 10 '14 at 13:08
  • take the case of $n=2$,the number is $0~mod 8$ for every m – Hawk Jan 10 '14 at 13:14
  • Why is $x^2 + 3 = 0 \pmod{3}$? What about $x=0$? Furthermore, that's not how division works in modulo. You likely want $ 3^{-1} \equiv 3 \pmod{8}$. – Calvin Lin Jan 10 '14 at 15:38
  • @CalvinLin Since $3n+1=x^2$ with $n$ positive integer how can $x=0$? Also $n+1=\frac{x^2+2}{3}$ isn't an integer? – Kal S. Jan 10 '14 at 16:42
  • Sorry, I screwed up the first part as I misread your statement. However, I still disagree with your division. It is possible that $\frac{x^2+2} { 3} \equiv 6 \pmod{8}$, which arises from $x^2 + 2 \equiv 2 \pmod{8}$. For example, when $x=4$. – Calvin Lin Jan 11 '14 at 03:00
  • @CalvinLin Now I think I understand what you mean. I skipped that part. I edited my answer. – Kal S. Jan 11 '14 at 10:09