You're reading an English translation of
the original Czech page.

In an effort to move closer to solutions of LF-theorem
could therefore make sense to know the rules for sums of Fermat coefficients.
Unlike the sequence of numbers n^{p} in a sequence of numbers f(n,p)
sums exist.
For given p we will write f(n,p) = f(n).

In case pεP the numbers n^{p} indicate total number
of instances in systems G(n,p).
This number is approximately p times the number of self classes
and numbers of self classes correspond to Fermat's coefficients.
f(n,p) = (n^{p}−n)/p.

For p=2:

(a,b,c) f(a)+f(b)= f(c) (a,b,c) f(a)+f(b)= f(c) ───────────────────────────── ─────────────────────────── ( 4, 6, 7) 6 + 15 = 21 ( 8, 28, 29) 28 +378 =406 ( 5, 10, 11) 10 + 45 = 55 ( 9, 11, 14) 36 + 55 = 91 ( 6, 7, 9) 15 + 21 = 36 ( 9, 36, 37) 36 +630 =666 ( 6, 15, 16) 15 +105 =120 ( 7, 10, 12) 21 + 45 = 66 ( 7, 21, 22) 21 +210 =231

For p=3: For p=5:

(a,b,c) f(a)+f(b)= f(c) (a,b,c) f(a)+f(b)= f(c) ─────────────────────────── ──────────────────────────── ( 9, 15, 16) 240+1120 =1360 (13,16,17) 74256+209712=283968 (21, 55, 56) 3080+55440=58520 (31, 56, 59) 9920+58520=68440

For higher values of Fermat coefficients it shows that to meet the equation f(a) + f(b) = f(c) both addends must be large enough. Values f(a), f(b) and therefore also a,b gradually flatten (to close levels). In the case of p = 2 for the solution (13, 78, 79); ie. 78 + 3003 = 3081, gives the ratio (a + b)/c the value of 1.15190.

Another solution is close to the value (a+b)/c = √2 = 2^{1/2}:

(1728,1768,2472) 1492128+1562028 =3054156 1,41424 (1738,1768,2479) 1509453+1562028 =3071481 1,41428 (1740,1794,2499) 1512930+1608321 =3121251 1,41417 (1751,1765,2486) 1532125+1556730 =3088855 1,41432

Similarly, in the case of p=3 ratio (a+b)/c grows towards the value 2^{2/3}.

( 923,1287,1429) 262109848+ 710581872 =972691720 1,54654 ( 969,1002,1242) 303284080+ 335337002 =638621082 1,58696 (1352,1479,1787) 823774952+1078407920 =190218287 1,58422

Generally, from the relationship 2∙f(a) = f(c) it follows:
a/c = 2^{1/p} tj.2a/c = 2^{(p−1)/p}.

We will call decisive sequence - the first difference sequence of a given arithmetical sequence. E.g. in the sequence of second powers it is the sequence R2 = {1,3,5,7,9,11,13,...} and in the sequence of third powers R3 = R3 = {1,7,19,37,61,91,127,.....} We are interested in what must be addends (of elements of decisive sequence), so that their sum is the k-th power of a number..

Eg. in sequence R3 (k=3):

Twoo addends: ------------- 3781+4219=8000 = 20^3 19+17557=17576 = 26^3 7651+25117=32768 = 11347+21421=32768 = 32^3 ...Numbers ...,20,26,32,... are of the form 2 + 6n = 2 + 2kn

Three addends: ------------- 7+91+631=37+61+631=61+271+397=91+169+469=127+271+331=...=729=9^3 7+1261+2107=19+919+2437=37+169+3169=37+547+2791=37+1387+1951=...=3375 = 15^3 19+331+8911=61+3781+5419=91+1519+7651=...=9261 =21^3Numbers ..,9,15,21,... are of the form 3 + 6n = 3 + 2kn

A similar result is obtained for 4, 5 or more addends. Generally, for p addends, the resulting power has the form:

This is simply the consequence of the fact that if a^2+b^k=c^k, k &eps; P then a+b-c = 2kq, q &eps; N.