Sue from Madras College first explored the factors of the numbers which are written as 10101 in different number bases, then proved that the numbers 10201, 11011 and 10101 are composite in any base, then she also proved the results that she had discovered. Here is her solution.
1. Written in base 10 we notice that the larger factor of the previous base becomes the smaller factor of the next base.
10101 _{2} = 7 x 3
10101 _{3} = 13 x 7
10101 _{4} = 21 x 13
10101 _{5} = 31 x 21
10101 _{6} = 43 x 31
10101 _{7} = 57 x 43
10101 _{8} = 73 x 57
10101 _{9} = 91 x 73
10101 = 111 x 91.
2. Here we see that one of the factors is always 111 (written in different bases) and the other factor has digits (the base number - 1) followed by 1.
In base 2: 10101 = 111 x 11
In base 3: 10101 = 111 x 21
In base 4: 10101 = 111 x 31
In base 5: 10101 = 111 x 41
In base 6: 10101 = 111 x 51
In base 7: 10101 = 111 x 61
In base 8: 10101 = 111 x 71
In base 9: 10101 = 111 x 81
In base 10: 10101 = 111 x 91.
3. Written in base 10, the factors form a number series 7, 13, 21, 31, 43, 57, 73, ...
4. The difference between the two factors is always twice the base number:
7-3 | = | 4 | = | 2x2 | |
13-7 | = | 6 | = | 2x3 | |
21-13 | = | 8 | = | 2x4 etc. |
10201 is composite in any base.
Proof The number 10201 in
base $a$ (for $a \geq3$) would be
10201 _{a} | = | 1 + 2 a ^{2} + a ^{4} |
= | (1 + a ^{2} ) ^{2} . |
So the number 10201 in any base a would always be the square number 101 ^{2} and hence it is always composite.
11011 is composite in any base.
Proof The
number 11011 can be written in base a as:
11011 _{a} | = | 1+ a + a ^{3} + a ^{4} |
= | (1 + a ^{3} )+( a + a ^{4} ) | |
= | (1 + a ^{3} )+ a (1 + a ^{3} ) | |
= | (1 + a )(1 + a ^{3} ) | |
= | (1+ a ) ^{2} (1- a + a ^{2} ). |
So 11011 always has the factors (1 + a ) ^{2} and (1 - a + a ^{2} ) so it is always composite.
10101 is composite in any base.
Proof To
prove 10101 is composite in any base we must factorise it.
l0l0l _{a} | = | a ^{4} + a ^{2} +1 |
= | ( a ^{2} +1) ^{2} - a ^{2} | |
= | [( a ^{2} +1)+ a ][( a ^{2} +1)- a ]. |
10101 _{a} therefore has the factors [ a ^{2} + a + 1] and [ a ^{2} - a + 1] and hence it is composite for all values of a .
Now we can explain what we found before:
1. In base ( a + 1) the smaller factor would be [( a + 1) ^{2} - ( a + 1) + 1] and this is equal to [ a ^{2} + a + 1] which is the larger factor in base a .
2. If 10101 _{a} = [ a ^{2} +
a + 1] [ a ^{2} - a + 1] it is
clear that
[ a ^{2} + a + 1] = 111
_{a} and
[ a ^{2} - a + 1] = a (
a - 1) + 1 = ( a - 1) l _{a}
3. The series 3, 7, 13, 21, 31, ... has the general formula n ^{2} + n + 1 ( n = base number).
4. If the two factors in base a are [ a ^{2} + a + 1] and [ a ^{2} - a + 1] then their difference would be 2 a which is what we observed.