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The only possibilities are $a = 2, \ b = 4$ and $a = 6, \ b = 4$.
With more digits the sums of squares are never bigger than some multiple of $7^2$. It can be proved that sequences cannot go on for ever without repeating patterns and so all sequences go into cycles or go to a fixed point. Moreover it can be proved that there are no fixed points with 3 or more digits.
Happy Octopus
Age 11 to 14
Challenge Level 
This is an investigation of sequences formed by writing down
the sum of the squares of the digits of each number to get the next
number in the sequence. A number is called happy if it starts a
sequence that goes to a number which is repeated over and over
again (e.g. 1, 1, 1, 1, $\ldots$) and this is called a fixed point.
Other numbers lead to cyclic patterns which repeat over and over
again. For example this sequence leads to a repeating
8-cycle:
25, 29, 85, 89, 145, 42, 20, 4, 16, 37, 58, 89, $\ldots$
Rachel Walker from the Mount School, York explained that in
base ten, 122 is a sad number because it has an 8-cycle like 25 and
this is her solution:
122, 9, 81, 65, 37, 58, 89, 145, 42, 20, 4, 16, 37,
58........
You can find a lot of 'happy numbers' such as: 310, 70, 86,
130, 7, 5555, 1212, 13, 1000 and other numbers which are variations
of these, e.g. swapping the numbers the other way around (13, 31)
and adding zeros onto the end of numbers (31, 310). Also you get
numbers that add up to other happy numbers e.g. 5555 - four fives
squared = 100.
The following sequences are the patterns of fixed points and
cycles in base eight that I have found. These are fixed points so
they are happy numbers and any numbers which start sequences ending
like this are happy numbers.
64, 64, 62, 64, 64, $\ldots$
24, 24, 24, 24, 24, $\ldots$
1, 1, 1, 1, ,1, $\ldots$
These are 2-cycles:
32, 15, 32, 15, 32, 15, $\ldots$
20, 4, 20, 4, 20, 4, $\ldots$
This is a 3-cycle:
31, 12, 5, 31, 12, 5, 31, 12, 5, $\ldots$
Pen Areecharoenlert, also from The Mount School, York found
some more happy numbers and cycles in base eight. Pen says "There
does not appear to be any pattern in which ones are happy, but it
looks as though there are 3 fixed points: 1, 24 and 64." This is
Pen's list.
1, 1, 1, $\ldots$ so 1 is
happy.
2, 4, 20, 4, 20, $\ldots$ so 2 is sad. (2-cycle)
3, 11, 2, 4, 20, $\ldots$ so 3 is sad. (2-cycle)
4 is sad. (2-cycle)
5, 31, 12, 5, $\ldots$ so 5 is sad. (3-cycle)
6, 44, 40, 20, 4, 20, $\ldots$ so 6 is sad. (2-cycle)
7, 61, 45, 51, 32, 15, 32, $\ldots$ so 7 is sad.
(2-cycle)
10, 1, 1, 1, 1$\ldots$ so 10 is
happy.
11, 2, 4, 20, 4, $\ldots$ so 11 is sad. (2-cycle)
12, 5, 31, 12, $\ldots$ so 12 is sad. (3-cycle)
13, 12, $\ldots$ so 13 is sad. (3-cycle)
14, 21, 5, $\ldots$ so 14 is sad. (3-cycle)
5, 32, 15,$\ldots$ so 15 is sad. (2-cycle)
16, 45, 51, 32, $\ldots$ so 16 is sad. (2-cycle)
17, 62, 50, 31,$\ldots$ so 17 is sad. (3-cycle)
20, 4, 20, 4, $\ldots$ so 20 is sad. (2-cycle)
21, 5, $\ldots$ so 21 is sad. (3-cycle)
22, 10, 1, 1,$\ldots$ so 22 is
happy.
23, 15, $\ldots$ so 23 is sad. (2-cycle)
24, 24, 24, 24, so 24 is
happy.
25, 35,, 42, 24, 24,$\ldots$ so 25
is happy.
26, 50, 31,$\ldots$ so 26 is sad. (3-cycle)
27, 65, 75, 112, 6, 44, 40, 20,$\ldots$ so 27 is sad.
(2-cycle)
30 , 11, 2, 4, 20, $\ldots$ so 30 is sad. (2-cycle)
31, 12, $\ldots$ so 31 is sad. (3-cycle)
32, 15, $\ldots$ so 32 is sad. (2-cycle)
33, 22, 10, 1, $\ldots$ so 33 is
happy.
34, 31, 12, $\ldots$ so 34 is sad. (3-cycle)
35, 42, 24, 24, $\ldots$ so 35 is
happy.
36, 55, 62, 50, 31, $\ldots$ so 36 is sad. (3-cycle)
37, 72, 65, 75, 112 so 37 is sad. (2-cycle)
40 > 20 > 4 > 20 so 40 is sad. . (2-cycle)
41, 21, 5, 31, 12, 5, $\ldots$ so 41 is sad. (3-cycle)
42, 24, 24,$\ldots$ so 42 is
happy.
43, 31, $\ldots$ so 43 is sad. (3-cycle)
44, 40, 20, 4, 20,$\ldots$ so 44 is sad. (2-cycle)
45, 51, 32, $\ldots$ so 45 is sad. (2-cycle)
46, 64, 64, 64, $\ldots$ so 46 is
happy.
47, 101, 2, 4, 20, $\ldots$ so 47 is sad. (2-cycle)
Claire Kruithof, Madras College, St Andrews proved that 24 and
64 are the only 2 digit fixed points in base 8. This is Claire's
proof:
"We wanted to see if there would be any more fixed points so
we used algebra. The number following $ab$ is $a^2 + b^2$ where $a$
and $b$ are between 0 and 7. For a fixed point we have:
\[a^2 + b^2 = 8a + b\] \[b^2 - b = 8a - a^2\] \[b(b - 1) = a
(8 - a)\]
Searching for possibilities
| B | a |
| 1 x 0 = 0 | 7 x 1 = 7 |
| 2 x 1 = 2 | 6 x 2 = 12 |
| 3 x 2 = 6 | 5 x 3 = 15 |
| 4 x 3 = 12 | 4 x 4 = 16 |
| 5 x 4 = 20 | |
| 6 x 5 = 30 | |
| 7 x 6 = 42 |
The only possibilities are $a = 2, \ b = 4$ and $a = 6, \ b = 4$.
With more digits the sums of squares are never bigger than some multiple of $7^2$. It can be proved that sequences cannot go on for ever without repeating patterns and so all sequences go into cycles or go to a fixed point. Moreover it can be proved that there are no fixed points with 3 or more digits.
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