Can you solve for divisibility?

Question

Problem:

Find the sum of all positive integers n such that $(1+2+..+n) \mid 15[(n+1)^2+(n+2)^2+..+(2n)^2]$.

My approach:

I have tried $\frac{n(n+1)}{2}$ instead of $(1+2+3+..+n)$. But that doesn't help. I have used some divisibility rules to solve it. But I failed . I tried to simplify R.H.S . That goes like this.

$15[(n+1)^2+(n+2)^2+..+(2n)^2]$

$\implies$ $15[(n^2+2n+1)+(n^2+4n+4)+(n^2+6n+9)+...+(n^2+2n^2+n^2)]$

$\implies 15[n\cdot n^2+(2n+4n+..+2n^2)+(1+4+9+..+n^2)]$

$\implies 15[n^3+2n\cdot (1+2+..+n)+(1^2+2^2+3^2+..+n^2)]$

$\implies 15[n^3 + 2n\cdot (1+2+..+n) + \frac{n\cdot (n+1)\cdot (2n+1)}{6}]$

$\implies 15[n^3 + 2n\cdot \frac{n\cdot (n+1)}{2} + \frac{n\cdot (n+1)\cdot (2n+1)}{6}]$

$\implies 15[n^3 + n^2\cdot (n+1) + \frac{n\cdot (n+1)\cdot (2n+1)}{6}]$

$\implies 15\cdot n^3 + 15\cdot n^2\cdot (n+1) + 5\cdot \frac{n\cdot (n+1)\cdot (2n+1)}{2}]$

But this doesn't work either.

Conclusion:

I am not able to solve it. I need some help. Any suggestions are appreciated. Feel free to help me.

Answer 1

Okay, so let's restate the problem. We need to find the sum of all positive integers $n$ such that:

$$ \frac{n(n+1)}{2} \mid 15\left[(n+1)^2 + (n+2)^2 + \dots + (2n)^2\right] $$

At first, we are going to simplify the expression. The sum of the first $ n $ integers is given by:

$$ S_1 = \frac{n(n+1)}{2} $$

The right-hand side, the sum of the squares from $ (n+1)$ to $ 2n$, can be expressed as:

$$ S_2 = (n+1)^2 + (n+2)^2 + \dots + (2n)^2 $$

To find $S_2$, we can use the sum of squares formula:

$$ \sum_{k=1}^m k^2 = \frac{m(m+1)(2m+1)}{6} $$

The sum of squares from $n+1$ to $2n$ is as follows:

$$ S_2 = \sum_{k=1}^{2n} k^2 - \sum_{k=1}^{n} k^2 = \frac{2n(2n+1)(4n+1)}{6} - \frac{n(n+1)(2n+1)}{6}, $$ since we are only looking at n+1 up to 2n numbers. Simplyfing leads to:

$$ S_2 = \frac{(2n+1)\left[2n(4n+1) - n(n+1)\right]}{6} $$

We are now rewriting the expression inside the brackets:

$$ 2n(4n+1) - n(n+1) = 8n^2 + 2n - n^2 - n = 7n^2 + n $$

Thus, $S_2$ becomes:

$$ S_2 = \frac{(2n+1)(7n^2 + n)}{6} = \frac{n(7n+1)(2n+1)}{6} $$

Since we can cross out our fractions,

$$ \frac{n(n+1)}{2} \mid 15 \times \frac{n(7n+1)(2n+1)}{6} $$

cancelling out $n$ on both sides (assuming $n \neq 0$) and multiply through by 2 leads to

$$ n+1 \mid 5 \times (7n+1) \times (2n+1). $$

Now, the key is to analyze when $n+1$ divides $5 \times (7n+1) \times (2n+1) $.

To find the sum of all such $ n $, we would consider the expression modulo $ n+1$, simplifying it further, and determining for which $ n $ the condition holds. At this stage, the problem reduces to evaluating the divisibility by solving the equation:

$$ n+1 \mid 5 \times \left[(7n+1)(2n+1)\right] \quad \text{mod} \quad (n+1) $$

The remaining task is to determine the exact values of $n $ by inspecting the divisibility condition carefully, considering cases modulo $n+1 $.

Answer 2

Why don't you simply combine your two results?

$1+2+3+...+n = \frac{n \cdot (n+1)}{2}$

$15[(n+1)^2+(n+2)^2+..+(2n)^2]=15\cdot n^3 + 15\cdot n^2\cdot (n+1) + 5\cdot \frac{n\cdot (n+1)\cdot (2n+1)}{2}]$

Hence:

$(1+2+..+n) \mid 5[(n+1)^2+(n+2)^2+..+(2n)^2$

becomes:

$\frac{n \cdot (n+1)}{2} | 15\cdot n^3 + 15\cdot n^2\cdot (n+1) + 5\cdot \frac{n\cdot (n+1)\cdot (2n+1)}{2}$

You already know that $\frac{n \cdot (n+1)}{2} | 15\cdot n^2\cdot (n+1)$.
You also already know that $\frac{n \cdot (n+1)}{2} | 5\cdot \frac{n\cdot (n+1)\cdot (2n+1)}{2}$.

So, you are left with $\frac{n \cdot (n+1)}{2} | 15n^3$.

Solve that and you're there.

Answer 3

You were almost there !

Starting from $\quad\frac 12 n(n+1)\,\mid\, 15n^3+15n^2(n+1)+\frac 52 n(n+1)(2n+1)$

Just eliminate the terms that are already factor of $\frac 12n(n+1)$

and it reduces to $\frac 12n(n+1)\,\mid\, 15n^3$

Which is equivalent to $(n+1)\,\mid\, 30n^2$ or $n=0$

Since $n=0$ is a trivial solution of the original problem ($0=0$), this is fine, or you can just reject it considering only strictly positive $n$.

Now just express $n^2=(n+1)^2-2n-1=(n+1)^2-2(n+1)+1$

And only the constant term remains in the divisibility criteria.

The condition then just reduces to $(n+1)\,\mid\, 30$ which is easy to solve.