A number theory problem by Ankush Tiwari

After struggling to solve this problem with pen and paper, I eventually resorted to writing a Python programme to "spoon-feed" me the answer. Since this question does not fall into any distinct category, I would presume that programming solutions are allowed.

Here's my the Python programme:

import math

def perfectsquare (n):  
    root = math.sqrt(n)
    if int(root)**2 == n:
        return True
    else:
        return False

possiblevalues = []
Sum = 0

for x in xrange(1, 1000):
    Sum = Sum + x**2

    if perfectsquare(Sum) == True:
        possiblevalues.append(x)

print max(possiblevalues)

The programme outputs 24 as the answer.

Sum the series as (1/6)n(n+1)(2n+1)=m^2

The solutions are n=24 and m=70, which gives answer as 24.

I used the following python code. Also I used the while True loop which is an infinite loop so that I could find other possibilities besides 24

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s=0
i=1
import math
while True:
    s=s+math.pow(i,2)
    i=i+1
    if math.sqrt(s)-int(math.sqrt(s))==0 and i != 2:
        print (i-1)

The answer in the question is wrong as 24 isn't the largest natural number such that the given sum is a perfect square

Output 24 2547750 2828093 3974004 4636697 5019548 5065968 5529976 6434771 6741437 7101496 7898119 8192485 9530917 9774922 9916100 10303942 10514246 10656671 11945152 12032855 12153127 12799512 12819261 13801485 15276205 15347381 15960792 16108784 16236906 16357744 16400624 16439442 16457969 16676437 16866476 17203662 17315619 17440708 17576703 17592831 17798035 17829491 17844733 17918157 18401354 18439817 18604362 18849433 18868909 18982624

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24

#include<iostream>
#include<math.h>
using namespace std;
int main()
{
    int arr[100000];
    long long int sum=0;
    int count=0;
    for(long long int i=1;i<100000;i++)
    {
        sum=sum+i*i;
        double x=sqrt(sum);
        long long int y=sqrt(sum);
        if(x==y)
        {
            arr[count]=i;
            count++;
        }
    }
    for(int i=0;i<count;i++)
    {
        cout<<arr[i]<<endl;
    }
} 

OUTPUT

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24