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: {{begin-eqn}} {{eqn | l = \map {\sigma_2} n       | r = \sum_{d \mathop \divides n} d^2 }} {{eqn | r = \sum_{d \mathop \divides n} \paren {\frac n d}^2 }} {{eqn | r = n^2 \sum_{d \mathop \divides n} \paren {\frac 1 d}^2 }} {{eqn | o = <       | r = n^2 \sum_{d \text{ odd} } \paren {\frac 1 d}^2       | c = since $n$ is odd }} {{eqn | r = \frac {n^2 \pi^2} 8       | c = Sum of Reciprocals of Squares of Odd Integers }} {{eqn | l = \map {\sigma_2} {n \pm 1}       | r = \sum_{d \mathop \divides n \pm 1} d^2 }} {{eqn | o = \ge       | r = 1^2 + 2^2 + \paren {\frac {n \pm 1} 2}^2 + \paren {n \pm 1}^2       | c = since $n \pm 1$ is even and $n - 1 > 4$ }} {{eqn | r = \frac 5 4 n^2 \pm \frac 5 2 n + \frac {15} 2 }} {{end-eqn}} Therefore: {{begin-eqn}} {{eqn | l = \map {\sigma_2} {n + 1}       | r = \frac 5 4 n^2 + \frac 5 2 n + \frac {15} 2 }} {{eqn | o = >       | r = \frac {\pi^2} 8 n^2 }} {{eqn | o = >       | r = \map {\sigma_2} n }} {{end-eqn}} 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: {{begin-eqn}} {{eqn | l = \map {\sigma_2} n       | r = \sum_{i \mathop = 0}^k p^{2 i}       | c = Divisors of Power of Prime }} {{eqn | r = n^2 \sum_{i \mathop = 0}^k \frac 1 {p^{2 i} } }} {{eqn | o = <       | r = n^2 \sum_{i \mathop = 0}^\infty \frac 1 {p^{2 i} } }} {{eqn | r = \frac {n^2} {1 - \frac 1 {p^2} }       | c = Sum of Infinite Geometric Sequence/Corollary 2 }} {{eqn | r = n^2 \paren {1 + \frac 1 {p^2 - 1} } }} {{eqn | o = \le       | r = n^2 \paren {1 + \frac 1 {3^2 - 1} } }} {{eqn | r = \frac 9 8 n^2 }} {{end-eqn}} We have: {{begin-eqn}} {{eqn | l = \map {\sigma_2} {n - 1} - \map {\sigma_2} n       | o = >       | r = \paren {\frac 5 4 - \frac 9 8} n^2 - \frac 5 2 n + \frac {15} 2 }} {{eqn | r = \frac 1 8 \paren {n^2 - 20 n + 60} }} {{eqn | r = \frac 1 8 \paren {n - 10}^2 - 5 }} {{end-eqn}} 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: {{begin-eqn}} {{eqn | l = \map {\sigma_2} n       | r = 1^2 + p^2 + q^2 + p^2 q^2       | c = Product of Two Distinct Primes has 4 Positive Divisors }} {{eqn | r = n^2 \paren {1 + \frac 1 {p^2} + \frac 1 {q^2} + \frac 1 {n^2} } }} {{eqn | o = \le       | r = n^2 \paren {1 + \frac 1 {3^2} + \frac 1 {5^2} + \frac 1 {15^2} } }} {{eqn | r = \frac {52} {45} n^2 }} {{eqn | o = <       | r = \frac 9 8 n^2 }} {{end-eqn}} We have: {{begin-eqn}} {{eqn | l = \map {\sigma_2} {n - 1} - \map {\sigma_2} n       | o = >       | r = \paren {\frac 5 4 - \frac 52 45} n^2 - \frac 5 2 n + \frac {15} 2 }} {{eqn | r = \frac {17} {180} \paren {n^2 - \frac {450} {17} n} + \frac {15} 2 }} {{eqn | r = \frac {17} {180} \paren {n - \frac {225} {17} }^2 - \frac {615} {68} }} {{end-eqn}} The above is greater than $0$ when $n \ge 25$. Therefore we just need to check the following $n \le 149$: :$3, 5, 7, 9, 11, 13, 15, 21, 45, 63, 75, 99, 105, 117, 135, 147$ {{begin-eqn}} {{eqn | l = \map {\sigma_2} 1       | r = 1^2 }} {{eqn | r = 1 }} {{eqn | l = \map {\sigma_2} 2       | r = 1^2 + 2^2 }} {{eqn | r = 5 }} {{eqn | l = \map {\sigma_2} 3       | r = 1^2 + 3^3 }} {{eqn | r = 10 }} {{eqn | l = \map {\sigma_2} 4       | r = 1^2 + 2^2 + 4^2 }} {{eqn | r = 21 }} {{eqn | l = \map {\sigma_2} 5       | r = 1^2 + 5^2 }} {{eqn | r = 26 }} {{eqn | l = \map {\sigma_2} 6       | r = 1^2 + 2^2 + 3^2 + 6^2 }} {{eqn | r = 50 }} {{eqn | l = \map {\sigma_2} 7       | r = 1^2 + 7^2 }} {{eqn | r = 50 }} {{eqn | l = \map {\sigma_2} 8       | r = 1^2 + 2^2 + 4^2 + 8^2 }} {{eqn | r = 85 }} {{eqn | l = \map {\sigma_2} 9       | r = 1^2 + 3^2 + 9^2 }} {{eqn | r = 91 }} {{eqn | l = \map {\sigma_2} {10}       | r = 1^2 + 2^2 + 5^2 + 10^2 }} {{eqn | r = 130 }} {{eqn | l = \map {\sigma_2} {11}       | r = 1^2 + 11^2 }} {{eqn | r = 122 }} {{eqn | l = \map {\sigma_2} {12}       | r = 1^2 + 2^2 + 3^2 + 4^2 + 6^2 + 12^2 }} {{eqn | r = 210 }} {{eqn | l = \map {\sigma_2} {13}       | r = 1^2 + 13^2 }} {{eqn | r = 170 }} {{eqn | l = \map {\sigma_2} {14}       | r = 1^2 + 2^2 + 7^2 + 14^2 }} {{eqn | r = 250 }} {{eqn | l = \map {\sigma_2} {15}       | r = 1^2 + 3^2 + 5^2 + 15^2 }} {{eqn | r = 260 }} {{eqn | l = \map {\sigma_2} {20}       | r = 1^2 + 2^2 + 4^2 + 5^2 + 10^2 + 20^2 }} {{eqn | r = 546 }} {{eqn | l = \map {\sigma_2} {21}       | r = 1^2 + 3^2 + 7^2 + 21^2 }} {{eqn | r = 500 }} {{eqn | l = \map {\sigma_2} {44}       | r = 1^2 + 2^2 + 4^2 + 11^2 + 22^2 + 44^2 }} {{eqn | r = 2562 }} {{eqn | l = \map {\sigma_2} {45}       | r = 1^2 + 3^2 + 5^2 + 9^2 + 15^2 + 45^2 }} {{eqn | r = 2366 }} {{eqn | l = \map {\sigma_2} {62}       | r = 1^2 + 2^2 + 31^2 + 62^2 }} {{eqn | r = 4810 }} {{eqn | l = \map {\sigma_2} {63}       | r = 1^2 + 3^2 + 7^2 + 9^2 + 21^2 + 63^2 }} {{eqn | r = 4550 }} {{eqn | l = \map {\sigma_2} {74}       | r = 1^2 + 2^2 + 37^2 + 74^2 }} {{eqn | r = 6850 }} {{eqn | l = \map {\sigma_2} {75}       | r = 1^2 + 3^2 + 5^2 + 15^2 + 25^2 + 75^2 }} {{eqn | r = 6510 }} {{eqn | l = \map {\sigma_2} {98}       | r = 1^2 + 2^2 + 7^2 + 14^2 + 49^2 + 98^2 }} {{eqn | r = 12255 }} {{eqn | l = \map {\sigma_2} {99}       | r = 1^2 + 3^2 + 9^2 + 11^2 + 33^2 + 99^2 }} {{eqn | r = 11102 }} {{eqn | l = \map {\sigma_2} {104}       | r = 1^2 + 2^2 + 4^2 + 8^2 + 13^2 + 26^2 + 52^2 + 104^2 }} {{eqn | r = 14450 }} {{eqn | l = \map {\sigma_2} {105}       | r = 1^2 + 3^2 + 5^2 + 7^2 + 15^2 + 21^2 + 35^2 + 105^2 }} {{eqn | r = 13000 }} {{eqn | l = \map {\sigma_2} {116}       | r = 1^2 + 2^2 + 4^2 + 29^2 + 58^2 + 116^2 }} {{eqn | r = 17682 }} {{eqn | l = \map {\sigma_2} {117}       | r = 1^2 + 3^2 + 9^2 + 13^2 + 39^2 + 117^2 }} {{eqn | r = 15470 }} {{eqn | l = \map {\sigma_2} {134}       | r = 1^2 + 2^2 + 67^2 + 134^2 }} {{eqn | r = 22450 }} {{eqn | l = \map {\sigma_2} {135}       | r = 1^2 + 3^2 + 5^2 + 9^2 + 15^2 + 27^2 + 45^2 + 135^2 }} {{eqn | r = 21320 }} {{eqn | l = \map {\sigma_2} {146}       | r = 1^2 + 2^2 + 73^2 + 146^2 }} {{eqn | r = 26650 }} {{eqn | l = \map {\sigma_2} {147}       | r = 1^2 + 3^2 + 7^2 + 21^2 + 49^2 + 147^2 }} {{eqn | r = 24510 }} {{end-eqn}} 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 $n \ge 8$. {{qed}}