Some Results on Analysis and number theory
B.M.Cerna Maguiña, Victor H. López Solís, Dik D. Lujerio Garcia
SSome Results on Number Theory and
Analysis
B. M. Cerna Magui˜na, Victor H. L´opez Sol´ıs andDik D. Lujerio GarciaDepartamento Acad´emico de Matem´aticaFacultad de CienciasUniversidad Nacional Santiago Ant´unez de MayoloCampus Shancay´an, Av. Centenario 200, Huaraz, Per´[email protected]@[email protected] 25, 2021
In memory of Emilia Magui˜na Cabana.
Abstract
In this paper we obtain bounds for integer solutions of quadraticpolynomials in two variables that represent a natural number. Alsowe get some results on twin prime numbers. In addition, we use linearfunctionals to prove some results of the mathematical analysis and theFermat’s last theorem.
Key Words : Quadratic polynomials in two variables, twin prime numbers,Fermat’s last theorem. a r X i v : . [ m a t h . G M ] F e b Introduction
In [1] appears a technique to find integer solutions of quadratic polynomialsin two variables, the lower bound found in that article was not the best.In this paper we were able to find a lower bound that helps us improvethe technique to obtain integer solutions of quadratic polynomials in twovariables that represent a natural number. We also get some results on twinprime numbers. We show through a Lemma that the mentioned techniquecan be used to obtain important results, as is the case in the proofs of theH¨older and Minkowski inequalities in which it is necessary to demonstratethe useful inequality a λ b u ≤ λa + ub with λ + u = 1 and a, b, λ, u are positivenumbers. Finally we prove Fermat’s last theorem (when n is odd) usingelementary calculus techniques. First consider the following general case about bounds.
Theorem 2.1.
Let P be a natural number ending in one. If there is ( A, B ) ∈ N × N such that: (i) P = (10 A + 9)(10 B + 9) or (ii) P = (10 A + 1)(10 B + 1) or (iii) P = (10 A + 7)(10 B + 3) . Then we havefor (i) √ P − ≤ A + B ≤ √ P − , A < B. for (ii) √ P − ≤ A + B ≤ √ P − , A < B. for (iii) √ P + 28 − ≤ A + B ≤ (cid:0) √ P + 2 − (cid:1) , A < B. roof. For equations P = (10 x + 9)(10 y + 9) and P = (10 x + 1)(10 y + 1) theproof process is the same.If ( A, B ) ∈ N × N is a solution of the equation (i), then ( B, A ) is also asolution of (i). The line L through the points ( A, B ) and (
B, A ) is L : x + y = A + B . The line L T through the point x = y = √ P − and furthermore, asthis line being tangent to the curve (i) is given by L T : x + y = √ P − (seeFigure 1). Figure 1: The curve L T . The proyections of the vector ( x, y ) ∈ L, L T when x = y and the vector( x , y ) on the x -axis and intersection of the line L T with the x -axis giventhe results.For the equation (iii) we have P = (10 A + 3)(10 B + 3) + 4(10 B + 3) , from this last relationship we have q = P − B + 3) = (10 A + 3)(10 B + 3) (1)3s it was done in (i) we apply it in (1) so √ q − ≤ A + B ≤
25 [ √ q −
3] (2)By (1), P = q + 40 B + 12, from this relationship and (2) we get B ≤ P − √ q − ≤
25 [ √ P − −
3] (3)Then by (1) and (3) we have √ q − ≥ (cid:113) P +245 − . (4)Also of the relations (2), (3) and (4) we have: (cid:114) P + 245 − ≤ A + B ≤ (cid:104) √ P − − (cid:105) . (5)Also the equation (1) can be written as P = (10 A + 7)(10 B + 7) − A + 7)from this relationship we have P + 4(10 A + 7) = (10 A + 7)(10 B + 7) . (6)Using relation (i) in (6) √ P + 28 + 40 A − ≤ A + B ≤ (cid:104) √ P + 28 + 40 A − (cid:105) (7)of relationships (5) and (7) with A < B we get √ P + 28 − ≤ A + B ≤ (cid:104) √ P − − (cid:105) . (8) Example 2.1.
In the theorem 2.1 of [1] we describe an example, where ) AB = 4500 + 4500 NM − − N M b) A + B = 5000 − NM − + N M c) AB + A + B = 9500 − NM − + N M with τ = 500 NM + − NM and τ asume posibles valores para A ≥ .From (b) and using (i) we get , ≤ NM ≤ , . (9) Hence replacing τ in (9), we have ≤ τ ≤ . In addition, the followingexpressions are obtained ( A − B −
1) = − τ ; ( A +1)( B +1) = 501 − τ ; AB = 9 τ +4499 (10) and from there, for any value of A = ˚3 + 2 or ˚3 + 1 we get τ = ˚3 . So τ ∈ { , , , , , } , τ assumes seven possible values.As P = ˚4 + 3 = 100 AB + 90( A + B ) + 81 = (10 A + 9)(10 B + 9) (11)˚4 + 3 = 2( A + B ) + 1 , thus A + B = ˚4 + 1 or A + B = ˚4 + 3 . Also by (11) we have A = ˚4 and B = ˚4 + 2 or A = ˚4 + 2 and B = ˚4 (12) and by (12) and (10) (2 A − B −
1) = − × τ (˚4 − B −
1) = 2 τ − (˚4 + 2)˚4 − B + 2 = 2 τ (13) where the result is equal when A = ˚4 + 2 y B = ˚4 .By (13) it is clear that τ = ˚4 + 1 or τ = ˚4 + 3 . Therefore τ ∈ { , , } .If A + B = 4 k + 1 , then from (10) we know that τ = ˚3 and by (b) A + B = 5001 − τ , so A + B = ˚3 , that is, A + B = 3 k = 4 k +1 = 5001 − τ ,which forces us to have τ = ˚6 which is false. Therefore A + B = 4 k + 3 (14)5 nd by (14) the following possibilities are obtained τ = 12 λ + 9 or τ = 12 λ + 3 (15) but A = 3 a + 1 , B = 3 b + 2 or A = 3 a + 2 , B = 3 a + 1 , replacing theserelations in (10) and (15) we get a (3 b + 1) = −
501 + 3(4 λ + 3) · and (3 a + 2)(3 b + 3) = 9501 − (12 λ + 9) . For A = 3 a + 1 , B = 3 b + 2 , A = ˚4 and B = ˚4 + 2 we get a = ˚4 and b = ˚4 or a = ˚4 + 1 and b = ˚4 + 3 , by replacing these values in (2.1) we get acontradiction.The same results are obtained for the other cases. Therefore, the onlypossibility that guarantees a solution is when τ = 12 λ + 3 that is τ ∈ { } . Using the ideas of the Theorem 2.1 of [1] we will demonstrate a veryimportant lemma which serves to demonstrate the H¨older inequality andconsequently the Minkowsky inequality.
Lemma 2.1. If a, b, λ and u are non-negative numbers and u + λ = 1 , then a λ b u ≤ λa + ub .Proof. Let F ( x, y ) = ( a λ b u ) x + ( λa + ub ) y be so it is clear that F is acontinuous linear functional, then Ker F = ( − ( λa + ub ) , a λ b u ), { Ker F } ⊥ =( a λ b u , λa + ub ). Thus F (1 ,
1) = a λ b u + λa + ub. (16)As R = Ker F ⊕ { Ker F } ⊥ we have(1 ,
1) = λ (cid:0) − ( λa + ub ) , a λ b u (cid:1) + λ (cid:0) a λ b u , λa + ub (cid:1) (17)and relationships (16) and (17) we get a λ b u + λa + ub = λ (cid:0) a λ b u + ( λa + ub ) (cid:1) (18)and a λ b u [1 − λ a λ b u ] = ( λa + ub ) [ λ ( λa + ub ) −
1] (19)which implies 1 − λ a λ b u ≥ λ ( λa + ub ) − ≥ − λ a λ b u ≤ λ ( λa + ub ) − ≤ . (21)By (20) we get λa + λb ≥ a λ b u (22)and by (21) λa + ub ≤ a λ b u . (23)We will show that (23) not happenes. By (23) and (19) λ (cid:2) a λ b u + λa + ub (cid:3) ≤ a, b (cid:54) = 0. Thus by (24) and using λ + u = 1 λ ≤ a if a ≤ b or λ ≤ b if b ≤ a (25)making a or b big enough, we get the only chance λ = 0.Therefore (23) is impossible, so only the relationship happens (22). Corollary 2.1.
Let H be a Pre-Hilbert space over C , then | < x, y > | ≤ √ < x, x > √ < y, y >. Now we will use the Theorem 2 . Theorem 2.2.
Let P be a natural number ending in and P = ˚3 + 2 .If (i) P = (10 x + 9)(10 y + 9) or (ii) P = (10 x + 1)(10 y + 1) or (iii) P = (10 x + 7)(10 y + 3) and (iv) P + 2 = (10 x + 9)(10 y + 7) or (v) P + 2 = (10 x + 1)(10 y + 3) . If there are integer solutions ( A, B ) ∈ N × N and ( C, D ) ∈ N × N of the quadratic equations representing P and P +2 , then (i) √ P − ≤ A + B ≤ √ P − or (ii) √ P − ≤ A + B ≤ √ P − or iii) √ P + 28 − ≤ A + B ≤ (cid:0) √ P − − (cid:1) with A < B and (iv) √ P + 20 − ≤ C + D ≤ (cid:0) √ P − − (cid:1) with C < D . or (v) √ P + 8 − ≤ C + D ≤ (cid:16) √ P − (cid:17) with C < D .Proof.
The results (i) , (ii) , (iii) , (iv) and (v) are obtained in a similar wayto what was done in the Theorem 2.1. Remark 2.1.
From the Theorem 2.2 is easy to obtain bounds for
A, B, C, D, AM and CD . To find the integer solutions it is necessary to use the Theorem 2.1of [1], where AB = ( P − M + N ) M , A + B = ( P − M − N ) MCD = ( P − M + N ) M , C + 9 D = ( P − M − N ) N . with A < B , C < D , N < M , M < N , N and M are relative primes, N and M are relative primes.Also for any A, B, C, D ∈ N we have A + B = ˚3 and C + D = ˚3 , and wecan use the technique of the Example 2.1. Theorem 2.3.
Let P be a natural number with P = ˚3 + 2 . If P and P + 2 are prime numbers, then exist m and n relatively prime such that p + 1 = m = √ n − and if n − p p where p and p are prime numbers, then P and P + 2 are prime numbers.Proof. Consider the aplication F ( x, y ) = P x +( P +2) y , similar to the Lemma2.1 we have Ker F = { ( − ( P + 2) , P ) } , { Ker F } ⊥ = { ( P, P + 2) } . Then F (1 ,
1) = P + P + 2 . (26)Also (1 ,
1) = λ ( − ( P + 2) , P ) + λ ( P, P + 2) (27)and by (26) and (27) we have P + 2 + P = λ (cid:0) P + ( P + 2) (cid:1) . (28)8ence λ ∈ Q . Let λ = mn be with m and n relatively prime. By (28) weget ( P + 2) ( m ( P + 2) − n ) = P ( n − mP ) (29)and using the fact that P + 2 and P are primes, then P + 1 = m = √ n − m = p + 1 = √ n −
1, then P + 2 p + 1 = n − P ( P + 2) = n − p p . Therefore P and P + 2 are prime numbers.In the following theorem we prove the Fermat’s last theorem using basictools developed in this article. Theorem 2.4.
If let
A, B, C ∈ N be are relatively prime numbers to eachother. For n ∈ N odd, n ≥ , the equation A n + B n = C n has no integersolution.Proof. Let F ( x, y, z ) = A n x + B n y + C n z be a real function of several vari-ables, then consider KerF = { ( − C n , , A n ) , ( − B n , A n , } , { KerF } ⊥ = { ( A n , B n , C n ) } . Supuse that equation A n + B n = C n (30)has integer solution, then F (1 , ,
1) = 2 C n . (31)Also (1 , ,
1) = λ ( − C n , , A n ) + λ ( − B n , A n ,
0) + λ ( A n , B n , C n ). Hence,applying F we obtain 2 C n = λ ( A n + B n + C n ) (32)From (32) A n = λ sin φ cos θB n = λ sin φ sin θC n = λ (cos φ + 1) (33)So by (33) and using (30) we get(cos θ + sin θ ) = (cos φ + 1)sin φ = K (34)9t is clear that θ = θ ( A, B, C, n ), φ = φ ( A, B, C, n ), λ = λ ( A, B, C, n ) and K = K ( A, B, C, n ), θ ∈ (cid:104) , π (cid:105) , φ ∈ (cid:104) , π (cid:105) .From (34) sin θ = (cid:114) − K √ − K θ = (cid:114) K √ − K φ = K − K + 1 and sin φ = 2 KK + 1 . Also from the third equation of (33) it followsthat cos φ ∈ Q because λ is rational number, so K ∈ Q , ≤ K ≤ √ θ ∈ Q making λK = ±√ − K with λ ∈ Q , then K = 2 λ + 1 and0 ≤ λ ≤ − ≤ λ ≤ λ and λ are rational numbers, consider λ = MN and λ = rt with G.C.D ( M, N ) = 1 and
G.C.D ( r, t ) = 1. As − ≤ λ ≤
1, using the relationsof (33), if we consider 0 ≤ λ ≤
1, we obtain C n = 4 N t M (3 t + r ) , B n = 2 tN( t − r ) M (3 t + r ) , A n = 2 N t ( t + r ) M (3 t + r ) . (35)Also, if we consider the case − ≤ λ ≤
0, we obtain analogous expressions.Without loss of generality, which will be supported later, from the equation(30), if we assume that C is even, then B and A are odd.We observe from (35) that if the factor 2 N tM (3 t + r ) is a natural num-ber greater than one, A, B and C have common prime factors, which is acontradiction. Then consider1 β = 2 N tM (3 t + r ) , β ∈ N , β ≥ . (36)From (36) and (35) we have C n = 1 β t, B n = 1 β ( t − r ) , A n = 1 β ( t + r ) . (37)If p (cid:54) = 2 was a prime divisor of the numbers β , t − r y t + r , then t − r = pθ , t + r = pθ and so 2 t = p ( θ + θ ), and as p (cid:54) = 2 implies that p divide to t , whichimplies that p divide to r , which is a contradiction, since G.C.D ( r, t ) = 1.Therefore, the unique prime divisor of β is 2, then C n = t, B n = t − r , A n = t + r C is even, t is even, so r is odd. Consequently, of (36)1 = 4 N tM (3 t + r ) (39)which implies that (38) y (39) are situations that will not happen. Therefore, β = 1 and so C n = 2 t, B n = t − r, A n = t + r. (40)If t was odd, (40) would be a contradiction, since 2 does not have an exactnth root and C ∈ I , which is a contradiction. Therefore, t is even and r isodd. Then t has the following form t = 2 n − p nα . . . p nα k k , n ≥ . (41)In addition from (40) we have B n − A n = 2 r . Now if n was 2, we would have( B − A )( B + A ) = 2 r , but B − A and B + A are pairs, which is absurd. Also,is easy to see that for the other even values of n the same contradiction isreached. Therefore n has to be odd.In addition of (36) we get2 N tM (3 t + r ) = 1 , for all N, M, t, r. (42)Now, we will justify why C was supposed to be even. If C is odd, then A iseven and B is odd. Define − A = (cid:98) A , − C = (cid:98) C , B = (cid:98) B . Thus, we consider G ( x, y, z ) = (cid:98) C n x + (cid:98) B n y + (cid:98) A n z, (cid:98) C n + (cid:98) B n = (cid:98) A n . Now, doing the same process done with the linear functional F , we get (cid:98) A n = 1 (cid:98) λ (cid:16) cos (cid:98) φ + 1 (cid:17) , (cid:98) B n = 1 (cid:98) λ sin (cid:98) θ sin (cid:98) φ, (cid:98) C n = 1 (cid:98) λ sin (cid:98) φ cos (cid:98) θ, (43)where (cid:98) θ ∈ (cid:104) π , π (cid:105) y (cid:98) φ ∈ (cid:104) , π (cid:105) . We also getcos (cid:98) φ + 1sin (cid:98) φ = cos (cid:98) θ + sin (cid:98) θ = K , K > A n = 2 t , B n = r − t , C n = r + t (44)11r A n = t , B n = r − t , C n = r + t K = 2 λ + 1 and λ = r t >
1, with
G.C.D ( r , t ) = 1. Therefore,(45) is a contradiction, because t is even and r is odd. Then it is justifiedthat this process is only valid if n is odd.From the relation (42) and the fact G.C.D ( M, N ) =
G.C.D ( t, r ) = 1, weconclude that N = 3 t + r and M = 2 t. (46)Thus, from the third equation of (33) C n = 2 NM cos φ . (47)Then by (47) and using the fact that C n = 2 t and by (46) we get t = √ N √ φ √ N (cid:112) cos φ + 1 . (48)So by (48) and (34) t = √ N √ K √ K + 1 , with K = 2 λ + 1 and λ = rt . (49)At once, by replacing the value of K we have t = √ K √ r + t . (50)If r + t was a perfect square, that is, if exist z ∈ N such that r + t = z , then K must be of the form K = √ αm such that G.C.D ( α, m ) = 1, with √ < αm ≤
1, becuase (1 ≤ K ≤ √ t = αm z (51)Therefore, it is clear that z = τ m for some τ ∈ N , so in r + t = z r + α τ = τ m , (52)12his implies that r and t have the common factor τ , which is a contradiction,because M.D.C ( t ; r ) = 1. Therefore, r + t cannot be a perfect square.Hence by (50), the expression of r + t should be as follows r + t = 2 K a w . (53)for some a, w ∈ N . Then replacing (53) in (50), we have t = √ K a +1 w √ wK a +1 (54)from there a = 2 b + 1 because K is a rational number, for some b ∈ N . So,we get t = wK b +1) , with K = 2 λ + 1 and λ = rt , the t = w. b +1 .t b +2 ( r + t ) b +1 , hence ( r + t ) b +1 = w · b +1 · t b +1 . This last relationship is a contradiction, since
G.C.D ( r, t ) = 1 and t is evenand r is odd.Therefore, this proves that the Fermat equation A n + B n = C n has nosolution if n is odd. Acknowledgment
The authors thank God for allowing this work to be carried out and com-pleted.The second author was supported and was funded by CONCYTEC-FONDECYT within the framework of the call “Proyecto Investigaci´on B´asica2019-01” [380-2019-FONDECYT].