2008 iTest Problems/Problem 83

Problem

Find the greatest natural number $n$ such that $n\leq 2008$ and $(1^2+2^2+3^2+\cdots + n^2)\left[(n+1)^2+(n+2)^2+(n+3)^2+\cdots + (2n)^2\right]$ is a perfect square.

Solution

Notice that $\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$ , so $\begin{align*} \sum_{i=n+1}^{2n} i^2 &= \sum_{i=1}^{2n} i^2 - \sum_{i=1}^n i^2 \\ &= \frac{2n(2n+1)(4n+1)}{6} - \frac{n(n+1)(2n+1)}{6} \\ &= \frac{16n^3 + 12n^2 + 2n}{6} - \frac{2n^3 + 3n^2 + n}{6} \\ &= \frac{14n^3 + 9n^2 + n}{6} \\ &= \frac{n(2n+1)(7n+1)}{6} \end{align*}$ Thus, $(1^2+2^2+3^2+\cdots + n^2)\left[(n+1)^2+(n+2)^2+(n+3)^2+\cdots + (2n)^2\right] = \frac{n^2 (2n+1)^2 (n+1)(7n+1)}{36}$ . In order for the expression to be a perfect square, $(n+1)(7n+1)$ must be a perfect square.

By using the Euclidean Algorithm, $\gcd(n+1,7n+1) = \gcd(n+1,6)$ . Thus, the GCD of $n+1$ and $7n+1$ must be factors of 6. Now, split the factors as different casework. Note that the quadratic residues of 7 are 0, 1, 2, and 4.

If $\gcd(n+1,7n+1) = 6$ , then $n \equiv 5 \pmod{6}$ . Let $n = 6a+5$ , so $(n+1)(7n+1) = (6a+6)(42a+36) = 36(a+1)(7a+6)$ . Since 6 is divided out of $n+1$ and $7n+1$ , $a+1$ and $7a+6$ are relatively prime, so $a+1$ and $7a+6$ must be perfect squares. However, since 6 is not a quadratic residue of 7, the GCD of $n+1$ and $7n+1$ can not be 6.
If $\gcd(n+1,7n+1) = 3$ , then $n \equiv 2 \pmod{3}$ . Let $n = 3a+2$ , so $(n+1)(7n+1) = (3a+3)(21a+15) = 9(a+1)(7a+5)$ . Since 3 is divided out of $n+1$ and $7n+1$ , $a+1$ and $7a+5$ are relatively prime, so $a+1$ and $7a+5$ must be perfect squares. However, since 5 is not a quadratic residue of 7, the GCD of $n+1$ and $7n+1$ can not be 3.
If $\gcd(n+1,7n+1) = 2$ , then $n \equiv 1 \pmod{2}$ . Let $n = 2a+1$ , so $(n+1)(7n+1) = (2a+2)(14a+8) = 4(a+1)(7a+4)$ . Since 2 is divided out of $n+1$ and $7n+1$ , $a+1$ and $7a+4$ are relatively prime, so $a+1$ and $7a+4$ must be perfect squares. We also know that $n+1$ and $7n+1$ do not share a factor of 3, so $n \equiv 1,3 \pmod{6}$ . That means $n \le 2007$ , so $a \le 1003$ . After trying values of $a$ that are one less than a perfect square, we find that the largest value that makes $(n+1)(7n+1)$ a perfect square is $a = 960$ . That means $n = 1921$ .
If $\gcd(n+1,7n+1) = 1$ , then $n+1 \equiv 1,5 \pmod{6}$ (to avoid common factors that are factors of 6), so $n \equiv 0,4 \pmod{6}$ . After trying values of $n$ that are one less than a perfect square, we find that the largest value that makes $(n+1)(7n+1)$ a perfect square is $n = 120$ (we could also stop searching once $n$ gets below 1921).

From the casework, the largest natural number $n$ that makes $(1^2+2^2+3^2+\cdots + n^2)\left[(n+1)^2+(n+2)^2+(n+3)^2+\cdots + (2n)^2\right]$ is a perfect square is $\boxed{1921}$ .

2008 iTest (Problems)
Preceded by: Problem 82	Followed by: Problem 84
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2008 iTest Problems/Problem 83

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