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Section 2. Mathematics
DOI: http://dx.doi.org/10.20534/AJT-17-1.2-23-25
Druzhinin Victor Vladimirovich, Holuskin Vladimir Semenovich, National research nuclear University «MEPhI», Sarov Institute of physics and technology, Department of mathematics, Sarov E-mail: vvdr@newmail.ru
Necessary conditions of the existence the sums of fermat
Abstract: The General necessary conditions for the solution of Diophantine equations, which includes the latest task of Fermat, formulated. The generalization of small Fermat's theorem is used. There are large number of examples.
Keywords: small and the last Fermat's theorem, comparison, deductions.
Under the sums of Fermat we understand Diophantine equation of the form:
F (n ;t ) = tpka'k = 0, (1)
k=1
where ff = ±1, ak and t natural number, t > 2. The Fermat's last theorem a' + b' - c' ^ 0 for t > 3 refers to the form (1). It is fully proven in 1995 by the British mathematician Edward Wiles [1], for which he in 2016 has the Abel prize. We consider the solution of equation (1) for an arbitrary number of summands and analyzed the necessary conditions of their existence. Trivial cases of subtraction of identical numbers or compensation for the expense of the pieces we reject. If these conditions are not met, then (1) in integers is not solved. If these conditions are met, then the evidence (1) requires additional analysis. Many such conditions are known, but we have found some new requirements. List them.
1. Signs f3k can't all be the same and odd numbers ak must be an even number.
2. If only one (ik = -1, then jjfikak > 0.
k=1
3. If t = p is a Prime number, due to the small Fermat's theorem a'k = ak (modt) and ^Pkak = Cp, where
k=1
C is an integer.
4. If we write a[ = 10akt 1 + akt0 in the decimal system, where akt0 > 0 the last digit of the num-
p-i p-i
ber EM. o =10C. If EM, o *10C (1) in integers is
not solved. Moreover, if we increase t to t + 4k, this amount will not change, i. e., with such increased amount of Fermat does not exist.
5. Choose randomly a Prime number p2 greater then t and all {ak} ± p2 (p2 mutually simple with ak). Generalization a small of Fermat's theorem in [2] for any t < p2 gives such a comparison:
a'k =n(ak ;t; p2 )(modp2), (2)
where /d(ak;t;p)deduction from the full set of minimum to module deduction, i. e. :
(P2 -1)/2 <v(ak ;t; p2 )<( -1)/2. The calculation of these deductions is relatively simple, and they have the following properties: Aak; P2-1; P2 ) = i;
; (p2 - 0/2; p2 ) = ±1; /d(i;t; p2 ) = 1;
V(p2 -l;i;p2) = (-!);
Aak ;2t i; p2 ) = v(p - ak ;2t i; P2);
^ (;2ti +1; p2 ) = -n(p2 - ak;2ti +1; p2);
§ v(ak;t;p2 )|<( P22 -1)M.
k=1
v(ak ;t; P2 ) = ^(ak ;t+p -1 Pi) ;
%(ak;(p2 -1)/2;p2) = 0.
k =1
As an example we give the matrix ofthe Fermat (tab. 1) the generalized small Fermat's theorem for p2 = 7.
(3)
k=1
k=1
Section 2. Mathematics
Table 1. - The matrix of Fermat. Deductions p for p2 = 7
a 1 2 3 4 5 6
t N,
6 1 1 1 1 1 1
5 1 -3 -2 2 3
4 1 2 -3 -3 2 1
3 1 1 -1 1 -1
2 1 -3 2 2 -3 1
1 1 2 3 -3 -2
Write the Fermat's amount F (n ;i) by applying the generalized small Fermat's theorem in the form:
F(n;t) = Jfika[ = F (n;t) + F2 (n;t), (4)
where:
ak = C kp2 + Mk;
F (n;t ) = P2 Z PC;
k=1
F2 (mt ) = Y pk ;t; P2).
(5)
From (4, 5) follows the fifth necessary condition in the form of a theorem: if F (n ;t ) = 0 ^ p2\ F2 (n;t) . If the Fermat's amount exists, F2 (n;i) divided by p2, including F2 may be zero. A special role plays the Central line t = (p2 -1)/2. We will continue to operate with two simpe numbers: the control p2 and the real t = pl. At the same p2 > pl. They are connected: p2 = pl +1; p2 = 2 pl +1; p2 = 4 pl +1; p2 = 8 pl +1 and so on.
If there is a control Prime number p2 = 2 p1 +1, all of {ak} ± p2, n is odd, then for 3 < n < p2 -2 F2 (n;pl) 0, i. e. in whole numbers the Fermat's amount does not exist. Give proof. Since F2 (n;pJ is the sum of an odd number of numbers «±1», then F2 (n; pl) 0
|F2(n;pl)\^(p2 -2) and the condition p2\F2(n;pt)
fails. Hence the quotient proof of the last Fermat's theorem. If {a;b;c} _L 7, then a3 + b3 -c3 ^ 0 as the amount
{±1; ±1; ±1} gives the number of {— 3;—1;1;3} is not a multiple of «7». Further, since 11 = 2 • 5 +1 and all { {ak} ± 11, the sum of three, five, seven, and nine numbers to the fifth power will never be equal to zero. In particular the calculation of F2 (n;pJ number of terms can be more (p2 -2). For example, consider the sum
F (5; 3 ) = (43 + 53 +103 +113 -163) .Then the residues modulo «7» are respectively equal to {l;-1;1;-1;l}, F2 (5;3) = -1, and the sum F (5;3) = -1576 = -7 • 225 -1, as it should be not equal to zero.
Let specific degrees. At t = 2 the control p2 = 3 . Hence, it follows that Pythagorean triple (type
32 + 42 - 52 = 0) must contain one number is a multiple of «3». If not, deductions of squares modulo «3»
3
equal to 1, and F2 (2;1) = = ±1 is not divisible by
k
«3». This requirement also follows from the quadratic identities:
4a2b2 +(a2 - b2 )2 =(a2 + b2)2. (6)
As shown by the author [3] for F (n ;2) = 0 we can obtain any length. For example:
32 + 42 +122 +842 + 36122 -36132 = 0. (7)
Here are only two numbers not divisible by «3»: {4;3613} and F2 (6;2) = 0.
Consider the degree t = 3. It is known that the triple of numbers (a3 + b3 - c3) 0 , and this equality was proved by Euler. However, the evidence is apparently so cumbersome that not a single textbook we never found him. Our method shortens the proof. Let a + b = c + 3T, T > 0, we must take a = 3al + l,b = 3bl + l,c = 3cl -1. If we build this equation in the cube, we get 1 = 3K, which is impossible. Now let a = 3al. In this case, we come to quadratic equation :
Tc2 + c (3T2 - ap) + 3T (T - abb) = 0.
The discriminant D = (3T2 + alb) - 12T4. According to the quadratic identity (5) to VD e N requires that T2 = alb. But this gives c = {0; -2T} ,which is unacceptable. So always (a3 + b3 - c3) ^ 0. If the number of sum-mands is more, there are chains of any length. We received the the Fermat's amount f forty-one components by the method of [4]:
2 • 23 + 33 + 53 + 2 • 43 + 2 • 63 + 73 +103 +123 +
+133 + 3 -143 + 2 • 153 + 4 463 +173 +193 + 243 +
(8)
+263 + 273 + 2 • 283 + 3 • 293 + 303 + 2 • 313 + 323 +
+353 + 363 + 383 + 393 + 623 - 923 = 0
Go to a degree t = 4. F (3;4) ^ 0 . This inequality has proved by Fermat himself. We it looks so. Let {a;b;c}_L 5, then a4 + b4 -c4 = 5T +1 at the second small Fermat's theorem. To the right of zero is impossible. Need to take, for example, a = 5ai. But there is superimposed a stronger necessary condition. We write a4 + b4 - c4 =(a2) +(b2 )2 -(c2 )2 = 0. From to (6) b2 =(a2 - fi2), c2 =(a2 + ^2). In integers the equation impossible. So, all a4 + b4 - c4 ^ 0 . Hence this inequality is in many other even t.
Euler put forward the hypothesis that the Quartet Quartets F (4; 4) ^ 0 for any integers. However, in recent years it was experimentally refuted. Computer calculations have a few two quartet quartets [5; 6]. For example:
k=1
9 5 8 004 + 2 1 75 1 94 + 4 1 45 604 - 42 2 4814 = 0, (9) 26824404 +153656392+ (10)
+187967604 - 206156734 = 0. Consider the necessary conditions for the existence of such quartets. If all four numbers ak 1 5, then F2 (4;4) = 3 -1 = 2, , and therefore F(4;4) * 0 . If only
4
one number is a multiple of «5», EA = {l^} and we
k
get the same result. Only when two numbers are multi-
4
ples of «5» = 0 and the equalities (9, 10) is pos-
k
sible. Offer the Fermat's amount of six terms: 24 + 24 + 34+ 44 + 44 -54 = 0.
6. There is also a sixth necessary condition for the existence of the Fermat's amounts. Before that we used or the top row of the Fermat's matrices with t = p — 1 or
the Central line at t = p. = (p2 -1)/2. Sure you can use
other strings. Let t = pl =(p2 -1) / 4. In this line deductions there are only four different numbers (±1; ±B) For example, when: p2 = 13, B = 5 ; p2 = 29, B = 12; p2 = 53, B = 30 . The result is easily calculated F2. For example, it is necessary to check F (5;13) .Can we of the five numbers {±1; ±30} to get the number «53»? It's impossible. Therefore, F2 (5;13) 53C and all amounts with ak 153 give F (5;13) * 0.
If we take t = pl =(p2 -1) / 8 then a row will have eight different deductions.
For example, if p2 = 41, t = 5 — a line contains deductions {±1; ±3 ± 9; ±14} . From the sum of the five numbers will receive a "41" is impossible. Therefore, if {a;b;c;d; f} _L 41 then a5 + 25 + c5+ d5 - f5 * 0.
References:
1. Wils A.//Annals of Mathematics. - 1995. - 141. - P. 443-551.
2. Druzhinin V. V.//NTVP - 2015. - № 5. - P. 22-24.
3. Druzhinin V. V.//NTVP - 2014. - № 1. - P. 19-21.
4. Druzhinin V. V., Holuskin V. S.//AJTNS - 2016. - № 4-5. - P. 3-4.
5. Graham Z., Knuth D., Patashnik O. Concrete mathematics. - Moscow:"Mir", 1998. - P. 313.
6. Science. № 18. Fermat. - Moskow: De Agostini, 2015. - P. 82.
