Consecutive Integers whose Sums of Squares of Divisors are Equal

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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:

\(\displaystyle \map {\sigma_2} n\) \(=\) \(\displaystyle \sum_{d \mathop \divides n} d^2\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{d \mathop \divides n} \paren {\frac n d}^2\)
\(\displaystyle \) \(=\) \(\displaystyle n^2 \sum_{d \mathop \divides n} \paren {\frac 1 d}^2\)
\(\displaystyle \) \(<\) \(\displaystyle n^2 \sum_{d \text{ odd} } \paren {\frac 1 d}^2\) since $n$ is odd
\(\displaystyle \) \(=\) \(\displaystyle \frac {n^2 \pi^2} 8\) Sum of Reciprocals of Squares of Odd Integers
\(\displaystyle \map {\sigma_2} {n \pm 1}\) \(=\) \(\displaystyle \sum_{d \mathop \divides n \pm 1} d^2\)
\(\displaystyle \) \(\ge\) \(\displaystyle 1^2 + 2^2 + \paren {\frac {n \pm 1} 2}^2 + \paren {n \pm 1}^2\) since $n \pm 1$ is even and $n - 1 > 4$
\(\displaystyle \) \(=\) \(\displaystyle \frac 5 4 n^2 \pm \frac 5 2 n + \frac {15} 2\)

Therefore:

\(\displaystyle \map {\sigma_2} {n + 1}\) \(=\) \(\displaystyle \frac 5 4 n^2 + \frac 5 2 n + \frac {15} 2\)
\(\displaystyle \) \(>\) \(\displaystyle \frac {\pi^2} 8 n^2\)
\(\displaystyle \) \(>\) \(\displaystyle \map {\sigma_2} n\)

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:

\(\displaystyle \map {\sigma_2} n\) \(=\) \(\displaystyle \sum_{i \mathop = 0}^k p^{2 i}\) Divisors of Power of Prime
\(\displaystyle \) \(=\) \(\displaystyle n^2 \sum_{i \mathop = 0}^k \frac 1 {p^{2 i} }\)
\(\displaystyle \) \(<\) \(\displaystyle n^2 \sum_{i \mathop = 0}^\infty \frac 1 {p^{2 i} }\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {n^2} {1 - \frac 1 {p^2} }\) Sum of Infinite Geometric Sequence/Corollary 2
\(\displaystyle \) \(=\) \(\displaystyle n^2 \paren {1 + \frac 1 {p^2 - 1} }\)
\(\displaystyle \) \(\le\) \(\displaystyle n^2 \paren {1 + \frac 1 {3^2 - 1} }\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 9 8 n^2\)

We have:

\(\displaystyle \map {\sigma_2} {n - 1} - \map {\sigma_2} n\) \(>\) \(\displaystyle \paren {\frac 5 4 - \frac 9 8} n^2 - \frac 5 2 n + \frac {15} 2\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 8 \paren {n^2 - 20 n + 60}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 8 \paren {n - 10}^2 - 5\)

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:

\(\displaystyle \map {\sigma_2} n\) \(=\) \(\displaystyle 1^2 + p^2 + q^2 + p^2 q^2\) Product of Two Distinct Primes has 4 Positive Divisors
\(\displaystyle \) \(=\) \(\displaystyle n^2 \paren {1 + \frac 1 {p^2} + \frac 1 {q^2} + \frac 1 {n^2} }\)
\(\displaystyle \) \(\le\) \(\displaystyle n^2 \paren {1 + \frac 1 {3^2} + \frac 1 {5^2} + \frac 1 {15^2} }\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {52} {45} n^2\)
\(\displaystyle \) \(<\) \(\displaystyle \frac 9 8 n^2\)

We have:

\(\displaystyle \map {\sigma_2} {n - 1} - \map {\sigma_2} n\) \(>\) \(\displaystyle \paren {\frac 5 4 - \frac 52 45} n^2 - \frac 5 2 n + \frac {15} 2\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {17} {180} \paren {n^2 - \frac {450} {17} n} + \frac {15} 2\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {17} {180} \paren {n - \frac {225} {17} }^2 - \frac {615} {68}\)

The above is greater than $0$ when $n \ge 25$.


Therefore we just need to check the following $n \le 149$:

$3, 5, 7, 9, 11, 13, 15, 21, 45, 63, 75, 99, 105, 117, 135, 147$
\(\displaystyle \map {\sigma_2} 1\) \(=\) \(\displaystyle 1^2\)
\(\displaystyle \) \(=\) \(\displaystyle 1\)
\(\displaystyle \map {\sigma_2} 2\) \(=\) \(\displaystyle 1^2 + 2^2\)
\(\displaystyle \) \(=\) \(\displaystyle 5\)
\(\displaystyle \map {\sigma_2} 3\) \(=\) \(\displaystyle 1^2 + 3^3\)
\(\displaystyle \) \(=\) \(\displaystyle 10\)
\(\displaystyle \map {\sigma_2} 4\) \(=\) \(\displaystyle 1^2 + 2^2 + 4^2\)
\(\displaystyle \) \(=\) \(\displaystyle 21\)
\(\displaystyle \map {\sigma_2} 5\) \(=\) \(\displaystyle 1^2 + 5^2\)
\(\displaystyle \) \(=\) \(\displaystyle 26\)
\(\displaystyle \map {\sigma_2} 6\) \(=\) \(\displaystyle 1^2 + 2^2 + 3^2 + 6^2\)
\(\displaystyle \) \(=\) \(\displaystyle 50\)
\(\displaystyle \map {\sigma_2} 7\) \(=\) \(\displaystyle 1^2 + 7^2\)
\(\displaystyle \) \(=\) \(\displaystyle 50\)
\(\displaystyle \map {\sigma_2} 8\) \(=\) \(\displaystyle 1^2 + 2^2 + 4^2 + 8^2\)
\(\displaystyle \) \(=\) \(\displaystyle 85\)
\(\displaystyle \map {\sigma_2} 9\) \(=\) \(\displaystyle 1^2 + 3^2 + 9^2\)
\(\displaystyle \) \(=\) \(\displaystyle 91\)
\(\displaystyle \map {\sigma_2} {10}\) \(=\) \(\displaystyle 1^2 + 2^2 + 5^2 + 10^2\)
\(\displaystyle \) \(=\) \(\displaystyle 130\)
\(\displaystyle \map {\sigma_2} {11}\) \(=\) \(\displaystyle 1^2 + 11^2\)
\(\displaystyle \) \(=\) \(\displaystyle 122\)
\(\displaystyle \map {\sigma_2} {12}\) \(=\) \(\displaystyle 1^2 + 2^2 + 3^2 + 4^2 + 6^2 + 12^2\)
\(\displaystyle \) \(=\) \(\displaystyle 210\)
\(\displaystyle \map {\sigma_2} {13}\) \(=\) \(\displaystyle 1^2 + 13^2\)
\(\displaystyle \) \(=\) \(\displaystyle 170\)
\(\displaystyle \map {\sigma_2} {14}\) \(=\) \(\displaystyle 1^2 + 2^2 + 7^2 + 14^2\)
\(\displaystyle \) \(=\) \(\displaystyle 250\)
\(\displaystyle \map {\sigma_2} {15}\) \(=\) \(\displaystyle 1^2 + 3^2 + 5^2 + 15^2\)
\(\displaystyle \) \(=\) \(\displaystyle 260\)
\(\displaystyle \map {\sigma_2} {20}\) \(=\) \(\displaystyle 1^2 + 2^2 + 4^2 + 5^2 + 10^2 + 20^2\)
\(\displaystyle \) \(=\) \(\displaystyle 546\)
\(\displaystyle \map {\sigma_2} {21}\) \(=\) \(\displaystyle 1^2 + 3^2 + 7^2 + 21^2\)
\(\displaystyle \) \(=\) \(\displaystyle 500\)
\(\displaystyle \map {\sigma_2} {44}\) \(=\) \(\displaystyle 1^2 + 2^2 + 4^2 + 11^2 + 22^2 + 44^2\)
\(\displaystyle \) \(=\) \(\displaystyle 2562\)
\(\displaystyle \map {\sigma_2} {45}\) \(=\) \(\displaystyle 1^2 + 3^2 + 5^2 + 9^2 + 15^2 + 45^2\)
\(\displaystyle \) \(=\) \(\displaystyle 2366\)
\(\displaystyle \map {\sigma_2} {62}\) \(=\) \(\displaystyle 1^2 + 2^2 + 31^2 + 62^2\)
\(\displaystyle \) \(=\) \(\displaystyle 4810\)
\(\displaystyle \map {\sigma_2} {63}\) \(=\) \(\displaystyle 1^2 + 3^2 + 7^2 + 9^2 + 21^2 + 63^2\)
\(\displaystyle \) \(=\) \(\displaystyle 4550\)
\(\displaystyle \map {\sigma_2} {74}\) \(=\) \(\displaystyle 1^2 + 2^2 + 37^2 + 74^2\)
\(\displaystyle \) \(=\) \(\displaystyle 6850\)
\(\displaystyle \map {\sigma_2} {75}\) \(=\) \(\displaystyle 1^2 + 3^2 + 5^2 + 15^2 + 25^2 + 75^2\)
\(\displaystyle \) \(=\) \(\displaystyle 6510\)
\(\displaystyle \map {\sigma_2} {98}\) \(=\) \(\displaystyle 1^2 + 2^2 + 7^2 + 14^2 + 49^2 + 98^2\)
\(\displaystyle \) \(=\) \(\displaystyle 12255\)
\(\displaystyle \map {\sigma_2} {99}\) \(=\) \(\displaystyle 1^2 + 3^2 + 9^2 + 11^2 + 33^2 + 99^2\)
\(\displaystyle \) \(=\) \(\displaystyle 11102\)
\(\displaystyle \map {\sigma_2} {104}\) \(=\) \(\displaystyle 1^2 + 2^2 + 4^2 + 8^2 + 13^2 + 26^2 + 52^2 + 104^2\)
\(\displaystyle \) \(=\) \(\displaystyle 14450\)
\(\displaystyle \map {\sigma_2} {105}\) \(=\) \(\displaystyle 1^2 + 3^2 + 5^2 + 7^2 + 15^2 + 21^2 + 35^2 + 105^2\)
\(\displaystyle \) \(=\) \(\displaystyle 13000\)
\(\displaystyle \map {\sigma_2} {116}\) \(=\) \(\displaystyle 1^2 + 2^2 + 4^2 + 29^2 + 58^2 + 116^2\)
\(\displaystyle \) \(=\) \(\displaystyle 17682\)
\(\displaystyle \map {\sigma_2} {117}\) \(=\) \(\displaystyle 1^2 + 3^2 + 9^2 + 13^2 + 39^2 + 117^2\)
\(\displaystyle \) \(=\) \(\displaystyle 15470\)
\(\displaystyle \map {\sigma_2} {134}\) \(=\) \(\displaystyle 1^2 + 2^2 + 67^2 + 134^2\)
\(\displaystyle \) \(=\) \(\displaystyle 22450\)
\(\displaystyle \map {\sigma_2} {135}\) \(=\) \(\displaystyle 1^2 + 3^2 + 5^2 + 9^2 + 15^2 + 27^2 + 45^2 + 135^2\)
\(\displaystyle \) \(=\) \(\displaystyle 21320\)
\(\displaystyle \map {\sigma_2} {146}\) \(=\) \(\displaystyle 1^2 + 2^2 + 73^2 + 146^2\)
\(\displaystyle \) \(=\) \(\displaystyle 26650\)
\(\displaystyle \map {\sigma_2} {147}\) \(=\) \(\displaystyle 1^2 + 3^2 + 7^2 + 21^2 + 49^2 + 147^2\)
\(\displaystyle \) \(=\) \(\displaystyle 24510\)

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