How do you know (I mean can you prove) that there is an integer solution to the equation?let, f=an+1,
so, 3an^2+3an+1=a1+a2+a3+.....+an1
choose randomly value for a1 to an1, check that you get a integer solution for an
10569  Number Theory
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Similar to what one & only pointed out, we can take advantage of cubic quadruples similar to pythagorean triples. The best quadruple that I have found is:
3 4 5 = 6 (exponents have been omitted for clarity)
This yields other quadrupals:
6 8 10 = 12
12 16 20 = 24
24 32 40 = 48
and so on, doubling the previous equation each time.
If you notice, these have a nice relationship, the previous answer equals the first number in the next line. We can use these equations and simply substitute until we fill the desired number of numbers.
If n is odd:
n = 3, 3 4 5 = 6
n = 5, 3 4 5 8 10 = 12
n = 7, 3 4 5 8 10 16 20 = 24
n = 9, 3 4 5 8 10 16 20 32 40 = 48
and so on.
If n is even start off with:
4 7 8 17 = 18 (base quintuple)
18 24 30 = 36 (base quadruple multiplied by 6)
36 48 60 = 72
72 96 120 = 144
and so on doubling the previous quadruple.
Using substitution:
n = 4, 4 7 8 17 = 18
n = 6, 4 7 8 17 (24 30 = 36)
n = 8, 4 7 8 17 24 30 (48 60 = 72)
n = 10, 4 7 8 17 24 30 48 60 (96 120 = 144)
and so on.
Two nice features of this algorithm are that it is O(n) and that you only need to use long 64bit integers since the numbers double every odd or even length and the length is at most 100.
LL Cool Jay
3 4 5 = 6 (exponents have been omitted for clarity)
This yields other quadrupals:
6 8 10 = 12
12 16 20 = 24
24 32 40 = 48
and so on, doubling the previous equation each time.
If you notice, these have a nice relationship, the previous answer equals the first number in the next line. We can use these equations and simply substitute until we fill the desired number of numbers.
If n is odd:
n = 3, 3 4 5 = 6
n = 5, 3 4 5 8 10 = 12
n = 7, 3 4 5 8 10 16 20 = 24
n = 9, 3 4 5 8 10 16 20 32 40 = 48
and so on.
If n is even start off with:
4 7 8 17 = 18 (base quintuple)
18 24 30 = 36 (base quadruple multiplied by 6)
36 48 60 = 72
72 96 120 = 144
and so on doubling the previous quadruple.
Using substitution:
n = 4, 4 7 8 17 = 18
n = 6, 4 7 8 17 (24 30 = 36)
n = 8, 4 7 8 17 24 30 (48 60 = 72)
n = 10, 4 7 8 17 24 30 48 60 (96 120 = 144)
and so on.
Two nice features of this algorithm are that it is O(n) and that you only need to use long 64bit integers since the numbers double every odd or even length and the length is at most 100.
LL Cool Jay
LL Cool Jay
The Formula Wizard
Jason Winokur
The Formula Wizard
Jason Winokur