Consecutive Integers whose Sums of Squares of Divisors are Equal
Theorem
The only two consecutive positive integers whose sums of the squares of their divisors are equal are $6$ and $7$.
Proof
The divisors of $6$ are
- $1, 2, 3, 6$
and so the sum of the squares of the divisors of $6$ is:
- $1^2 + 2^2 + 3^2 + 6^2 = 1 + 4 + 9 + 36 = 50$
The divisors of $7$ are
- $1, 7$
and so the sum of the squares of the divisors of $7$ is:
- $1^2 + 7^2 = 1 + 49 = 50$
It remains to be shown that there are no more.
Let $n \ge 7$ be an odd number.
Then both $n - 1$ and $n + 1$ are even.
Denote $\map {\sigma_2} n$ the sum of squares of the divisors of $n$.
We will show that:
- $\map {\sigma_2} {n + 1} > \map {\sigma_2} n$
- $\map {\sigma_2} {n - 1} > \map {\sigma_2} n$ for $n \ge 151$
Since:
\(\ds \map {\sigma_2} n\) | \(=\) | \(\ds \sum_{d \mathop \divides n} d^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{d \mathop \divides n} \paren {\frac n d}^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds n^2 \sum_{d \mathop \divides n} \paren {\frac 1 d}^2\) | ||||||||||||
\(\ds \) | \(<\) | \(\ds n^2 \sum_{d \text{ odd} } \paren {\frac 1 d}^2\) | since $n$ is odd | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {n^2 \pi^2} 8\) | Sum of Reciprocals of Squares of Odd Integers | |||||||||||
\(\ds \map {\sigma_2} {n \pm 1}\) | \(=\) | \(\ds \sum_{d \mathop \divides n \pm 1} d^2\) | ||||||||||||
\(\ds \) | \(\ge\) | \(\ds 1^2 + 2^2 + \paren {\frac {n \pm 1} 2}^2 + \paren {n \pm 1}^2\) | since $n \pm 1$ is even and $n - 1 > 4$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 5 4 n^2 \pm \frac 5 2 n + \frac {15} 2\) |
Therefore:
\(\ds \map {\sigma_2} {n + 1}\) | \(=\) | \(\ds \frac 5 4 n^2 + \frac 5 2 n + \frac {15} 2\) | ||||||||||||
\(\ds \) | \(>\) | \(\ds \frac {\pi^2} 8 n^2\) | ||||||||||||
\(\ds \) | \(>\) | \(\ds \map {\sigma_2} n\) |
and for $n - 1$:
- $\map {\sigma_2} {n - 1} - \map {\sigma_2} n = \dfrac {10 - \pi^2} 8 n^2 - \dfrac 5 2 n + \dfrac {15} 2$
By Solution to Quadratic Equation, the above is greater than zero when:
- $n > \dfrac {\paren {5/2} + \sqrt {\paren {5/2}^2 - 4 \paren {\paren {10 - \pi^2} / 8} \paren {15/2} } } {2 \paren {\paren {10 - \pi^2} / 8} } \approx 150.3$
hence there are no solutions for $\map {\sigma_2} {n - 1} = \map {\sigma_2} n$ for $n \ge 151$.
Our estimate of $\map {\sigma_2} n$ is very rough.
If $n$ is one of the following, we can get sharper estimates:
Suppose $n = p^k$ for a prime $p \ge 3$ and $k \ge 1$.
Then:
\(\ds \map {\sigma_2} n\) | \(=\) | \(\ds \sum_{i \mathop = 0}^k p^{2 i}\) | Divisors of Power of Prime | |||||||||||
\(\ds \) | \(=\) | \(\ds n^2 \sum_{i \mathop = 0}^k \frac 1 {p^{2 i} }\) | ||||||||||||
\(\ds \) | \(<\) | \(\ds n^2 \sum_{i \mathop = 0}^\infty \frac 1 {p^{2 i} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {n^2} {1 - \frac 1 {p^2} }\) | Sum of Infinite Geometric Sequence/Corollary 2 | |||||||||||
\(\ds \) | \(=\) | \(\ds n^2 \paren {1 + \frac 1 {p^2 - 1} }\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds n^2 \paren {1 + \frac 1 {3^2 - 1} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 9 8 n^2\) |
We have:
\(\ds \map {\sigma_2} {n - 1} - \map {\sigma_2} n\) | \(>\) | \(\ds \paren {\frac 5 4 - \frac 9 8} n^2 - \frac 5 2 n + \frac {15} 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 8 \paren {n^2 - 20 n + 60}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 8 \paren {n - 10}^2 - 5\) |
The above is greater than $0$ when $n \ge 17$.
Suppose $n = p q$, where $p, q \ge 3$ are primes and $p \ne q$.
Then:
\(\ds \map {\sigma_2} n\) | \(=\) | \(\ds 1^2 + p^2 + q^2 + p^2 q^2\) | Product of Two Distinct Primes has 4 Positive Divisors | |||||||||||
\(\ds \) | \(=\) | \(\ds n^2 \paren {1 + \frac 1 {p^2} + \frac 1 {q^2} + \frac 1 {n^2} }\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds n^2 \paren {1 + \frac 1 {3^2} + \frac 1 {5^2} + \frac 1 {15^2} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {52} {45} n^2\) | ||||||||||||
\(\ds \) | \(<\) | \(\ds \frac 9 8 n^2\) |
We have:
\(\ds \map {\sigma_2} {n - 1} - \map {\sigma_2} n\) | \(>\) | \(\ds \paren {\frac 5 4 - \frac 52 45} n^2 - \frac 5 2 n + \frac {15} 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {17} {180} \paren {n^2 - \frac {450} {17} n} + \frac {15} 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {17} {180} \paren {n - \frac {225} {17} }^2 - \frac {615} {68}\) |
The above is greater than $0$ when $n \ge 25$.
Therefore we just need to check the following $n \le 149$:
- $3, 5, 7, 9, 11, 13, 15, 21, 45, 63, 75, 99, 105, 117, 135, 147$
\(\ds \map {\sigma_2} 1\) | \(=\) | \(\ds 1^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1\) | ||||||||||||
\(\ds \map {\sigma_2} 2\) | \(=\) | \(\ds 1^2 + 2^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 5\) | ||||||||||||
\(\ds \map {\sigma_2} 3\) | \(=\) | \(\ds 1^2 + 3^3\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 10\) | ||||||||||||
\(\ds \map {\sigma_2} 4\) | \(=\) | \(\ds 1^2 + 2^2 + 4^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 21\) | ||||||||||||
\(\ds \map {\sigma_2} 5\) | \(=\) | \(\ds 1^2 + 5^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 26\) | ||||||||||||
\(\ds \map {\sigma_2} 6\) | \(=\) | \(\ds 1^2 + 2^2 + 3^2 + 6^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 50\) | ||||||||||||
\(\ds \map {\sigma_2} 7\) | \(=\) | \(\ds 1^2 + 7^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 50\) | ||||||||||||
\(\ds \map {\sigma_2} 8\) | \(=\) | \(\ds 1^2 + 2^2 + 4^2 + 8^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 85\) | ||||||||||||
\(\ds \map {\sigma_2} 9\) | \(=\) | \(\ds 1^2 + 3^2 + 9^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 91\) | ||||||||||||
\(\ds \map {\sigma_2} {10}\) | \(=\) | \(\ds 1^2 + 2^2 + 5^2 + 10^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 130\) | ||||||||||||
\(\ds \map {\sigma_2} {11}\) | \(=\) | \(\ds 1^2 + 11^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 122\) | ||||||||||||
\(\ds \map {\sigma_2} {12}\) | \(=\) | \(\ds 1^2 + 2^2 + 3^2 + 4^2 + 6^2 + 12^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 210\) | ||||||||||||
\(\ds \map {\sigma_2} {13}\) | \(=\) | \(\ds 1^2 + 13^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 170\) | ||||||||||||
\(\ds \map {\sigma_2} {14}\) | \(=\) | \(\ds 1^2 + 2^2 + 7^2 + 14^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 250\) | ||||||||||||
\(\ds \map {\sigma_2} {15}\) | \(=\) | \(\ds 1^2 + 3^2 + 5^2 + 15^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 260\) | ||||||||||||
\(\ds \map {\sigma_2} {20}\) | \(=\) | \(\ds 1^2 + 2^2 + 4^2 + 5^2 + 10^2 + 20^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 546\) | ||||||||||||
\(\ds \map {\sigma_2} {21}\) | \(=\) | \(\ds 1^2 + 3^2 + 7^2 + 21^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 500\) | ||||||||||||
\(\ds \map {\sigma_2} {44}\) | \(=\) | \(\ds 1^2 + 2^2 + 4^2 + 11^2 + 22^2 + 44^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2562\) | ||||||||||||
\(\ds \map {\sigma_2} {45}\) | \(=\) | \(\ds 1^2 + 3^2 + 5^2 + 9^2 + 15^2 + 45^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2366\) | ||||||||||||
\(\ds \map {\sigma_2} {62}\) | \(=\) | \(\ds 1^2 + 2^2 + 31^2 + 62^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 4810\) | ||||||||||||
\(\ds \map {\sigma_2} {63}\) | \(=\) | \(\ds 1^2 + 3^2 + 7^2 + 9^2 + 21^2 + 63^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 4550\) | ||||||||||||
\(\ds \map {\sigma_2} {74}\) | \(=\) | \(\ds 1^2 + 2^2 + 37^2 + 74^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 6850\) | ||||||||||||
\(\ds \map {\sigma_2} {75}\) | \(=\) | \(\ds 1^2 + 3^2 + 5^2 + 15^2 + 25^2 + 75^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 6510\) | ||||||||||||
\(\ds \map {\sigma_2} {98}\) | \(=\) | \(\ds 1^2 + 2^2 + 7^2 + 14^2 + 49^2 + 98^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 12255\) | ||||||||||||
\(\ds \map {\sigma_2} {99}\) | \(=\) | \(\ds 1^2 + 3^2 + 9^2 + 11^2 + 33^2 + 99^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 11102\) | ||||||||||||
\(\ds \map {\sigma_2} {104}\) | \(=\) | \(\ds 1^2 + 2^2 + 4^2 + 8^2 + 13^2 + 26^2 + 52^2 + 104^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 14450\) | ||||||||||||
\(\ds \map {\sigma_2} {105}\) | \(=\) | \(\ds 1^2 + 3^2 + 5^2 + 7^2 + 15^2 + 21^2 + 35^2 + 105^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 13000\) | ||||||||||||
\(\ds \map {\sigma_2} {116}\) | \(=\) | \(\ds 1^2 + 2^2 + 4^2 + 29^2 + 58^2 + 116^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 17682\) | ||||||||||||
\(\ds \map {\sigma_2} {117}\) | \(=\) | \(\ds 1^2 + 3^2 + 9^2 + 13^2 + 39^2 + 117^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 15470\) | ||||||||||||
\(\ds \map {\sigma_2} {134}\) | \(=\) | \(\ds 1^2 + 2^2 + 67^2 + 134^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 22450\) | ||||||||||||
\(\ds \map {\sigma_2} {135}\) | \(=\) | \(\ds 1^2 + 3^2 + 5^2 + 9^2 + 15^2 + 27^2 + 45^2 + 135^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 21320\) | ||||||||||||
\(\ds \map {\sigma_2} {146}\) | \(=\) | \(\ds 1^2 + 2^2 + 73^2 + 146^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 26650\) | ||||||||||||
\(\ds \map {\sigma_2} {147}\) | \(=\) | \(\ds 1^2 + 3^2 + 7^2 + 21^2 + 49^2 + 147^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 24510\) |
and thus the only pair is $\map {\sigma_2} 6 = \map {\sigma_2} 7 = 50$.
We have also inadvertently proved that $\map {\sigma_2} {2 n} > \map {\sigma_2} {2 n + 1}$ for $n \ge 8$.
$\blacksquare$
Sources
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $6$