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Problem :

find the smallest natural number $n>1$ such that :

$$n(n+1)(2n+1)=6m^{2}~~,m\in\mathbb{N}$$

I don't know the method to find $n$ but I try many time

$$n=1,2,3,....,24$$

Then I find $n=24$

$$24.25.49=6.70^{2}$$

Another try :

$$n\equiv r\pmod{6}~~,r=1,..,5$$

$n=6k$ then $k(6k+1)(12k+1)=m^{2}$

So $k,6k+1,12k+1$ is perfect square in same time I find $k=4$ so $n=24$ Also for all $n=6k+r$ I can't find $n$ ?

So I'm going to see you solution Thanks!

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Ellen Ellen
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  • If you are asking for the complete solution, this isn't easy. https://www.math.ubc.ca/~bennett/paper21.pdf – saulspatz Mar 01 '20 at 17:30
  • @saulspatz yes sir very difficult to me – Ellen Ellen Mar 01 '20 at 17:32
  • @JoséCarlosSantos How we can solve it with simple method ? – Ellen Ellen Mar 01 '20 at 17:48
  • I am not aware of any such solution. – José Carlos Santos Mar 01 '20 at 17:50
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    Well, you've found the smallest $n$ that works. Proving that there is no other $n$ is another matter. It has a long history, as described at the beginning of the paper I linked. There is an elementary solution at https://www.tandfonline.com/doi/pdf/10.1080/00029890.1990.11995558?needAccess=true but it isn't free. You should be able to find it in a math library, though. – saulspatz Mar 01 '20 at 17:51
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    This is Lucas' square pyramid problem :

    https://en.wikipedia.org/wiki/Cannonball_problem

    solved by Watson more than 100 years ago. There are no further solutions in positive integers. Check this post https://math.stackexchange.com/q/3491396/647719

     – Pruthviraj Mar 01 '20 at 21:20

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