700 Followers prob!

Number Theory Level 4

Let $f(x) = \sqrt{1^{2} + 2^{2} + 3^{2} + \cdots + (x-1)^{2} + x^{2}}$ .

Find the minimum value of $n \geq2$ such that $f(n)$ is a positive integer.

$n$ is a positive integer.

Clarification : $f(x)$ denote the square root of the sum of the squares of first $x$ positive integers.

The answer is 24.

2 solutions

Devendra Kumar Singh
Oct 25, 2015

Shashank Rammoorthy
Jul 21, 2015

We are given that

The sum of the squares of natural numbers up to n is $\frac { n(n+1)(2n+1) }{ 6 }$ This fact can be proved easily with induction.

Any square number is equivalent to 0 or 1 (mod 4). And we require our sum to be a square number.

Therefore, taking the first case, $\frac { n(n+1)(2n+1) }{ 6 } \equiv \quad 0\quad (mod\quad 4)$ The only way this is possible is if $n\quad \equiv \quad 0\quad (mod\quad 4)$ or alternatively, 3 mod 4. We also notice that the denominator of the fraction is 6, and therefore the number we are looking for ( $n(n+1)(2n+1)$ ) should be a multiple of 6.

By replacing n with numbers that are equivalent to 0 mod 4, we find that n= 24.

Moderator note:

The second part of your solution seems redundant. Note that $n(n+1)(2n+1)$ is always going to be a multiple of 6. It does not imply that $n$ must be a multiple of 6.

It seems to me that what you've done is gotten lucky that the smallest answer is of the form $0 \pmod{4}$ , and then you tried sufficient cases to get to the answer.

How can one classify all such solutions? (If at all possible).

700 Followers prob!

The answer is 24.

2 solutions

Moderator note:

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