A GENERALIZATION OF EULER S THEOREM ON CONGRUENCIES Florentin Smarandache University of New Mexico 00 College Road Gallup, NM 8730, USA E-mail: smarand@unmedu In the paragraphs which follow we will prove a result which replaces the theorem of Euler: If (a,m) =, then a ϕ( m) (mod m)", for the case when a and m are not relatively primes A Introductory concepts One supposes that m > 0 This assumption will not affect the generalization, because Euler s indicator satisfies the equality: ϕ ( m) =ϕ( m)(see []), and that the congruencies verify the following property: a b(mod m) a b( mod ( m) ) (see [] pp -3) In the case of congruence modulo 0, there is the relation of equality One denotes (a,b) the greatest common factor of the two integers a and b, and one chooses (a,b) > 0 B Lemmas, theorem Lemma : Let be a an integer and m a natural number > 0 There exist, m 0 from N such that a = a 0, m = m 0 and (a 0 ) = It is sufficient to choose = (a,m) In accordance with the definition of the greatest common factor (GCF), the quotients of a 0 and m 0 and of a and m by their GCF are relatively primes (see [3], pp 5-6) Lemma : With the notations of lemma, if and if: = d 0 d, m 0 = m d, (d 0,m ) = and d, then > d and m 0 > m, and if = d, then after a limited number of steps i one has > d i+ = (d i,m i ) a = a 0 ; (a 0 ) = (0) m = m 0 ;
= d 0 d ; (d 0,m ) = () m 0 = m d ; d From (0) and from () it results that a = a 0 = a 0 d 0 d therefore = d 0 d thus > d if d 0 From m 0 = m d we deduct that m 0 > m If = d then m 0 = m = k d z 0 ( z N * and k ) Therefore m = k d z 0 ; d = (d,m ) = (,k d z 0 ) After i = z steps, it results d i+ = (,k) < Lemma 3: For each integer a and for each natural number m > 0 one can build the following sequence of relations: a = a 0 ; (a 0 ) = (0) m = m 0 ; = d 0 d ; (d 0,m ) = () m 0 = m d ; d = ; ( ) = (s ) m s = m s ; = ; ( ) = (s) m s = m s ; One can build this sequence by applying lemma The sequence is limited, according to lemma, because after r steps, one has d > d and m > m, and 0 r 0 r after r steps, one has dr > d r r + and mr > m r r +, etc, and the m i are natural numbers One arrives at = because if one will construct again a limited number of relations ( s+ ),,( s+ r) with +r < Theorem: Let us have am Z, and m 0 Then a ϕ( m s ) + s a s (mod m) where s and m s are the same ones as in the lemmas above
Similar with the method followed previously, one can suppose m > 0 without reducing the generality From the sequence of relations from lemma 3, it results that: (0) () () (3) (s) a = a 0 = a 0 d 0 d = a 0 d 0 d d = = a 0 d 0 d and (0) () () (3) (s) m = m 0 = m d = m d d = = m s d and m s d = d m s From (0) it results that = (a,m), and from (i) that d i = (d i,m i ), for all i from {,,,s} = d 0 d d d = dd ds ds d = d d s s s = Therefore d d = (d 0 ) (d ) (d ) 3 ( ) s (d s ) s+ = (d 0 ) (d ) (d ) 3 ( ) s because = Thus m = (d 0 ) (d ) (d ) 3 ( ) s m s ; therefore m s m ; (s) (s) ( ) = (, m s ) and ( ds, ms) = (s-) = ( ) = ( ) therefore ( ds, ms) = (s-) = ( ds 3, ms ) = ( ds 3, ms ds ) = ( ds 3, msdsds ) therefore ( ds 3, ms) = (i+) = (d i,m i+ ) = (d i,m i+ d i+ ) = (d i,m i+ 3 d i+ 3 d i+ ) = = = (d i d i+ ) thus (d i ) =, and this is for all i from 0,,,s { } (0) = (a 0 ) = (a 0,d m s ) thus (a 0 ) = From the Euler s theorem results that: (d i ) ϕ( m s ) (mod m s ) for all i from 0,,,s { }, ϕ( m s ) a0 (mod m s ) 3
ϕ( m but a s ) ϕ( m 0 = a s ) 0 (d 0 ) ϕ(m s ) (d ) ϕ(m s ) ( ) ϕ( m s ) ϕ( m therefore s ) a (mod 3 ms ) s+ times ϕ( m s ) a (mod m s ) a s 0 (d 0 ) s (d ) s (d ) s 3 ( ) a ϕ(m s ) a s 0 (d 0 ) s (d ) s ( ) (mod m s ) Multiplying by: (d 0 ) (d ) (d ) 3 ( ) s ( ) s we obtain: s s s s s ms a0( d0) ( d) ( ds ) ( ds ) a ϕ( ) s s s s s s a ( d ) ( d ) ( d ) ( d ) (mod( d ) ( d ) m ) 0 0 s s 0 s s but a s 0 (d 0 ) s (d ) s ( ) s a ϕ(m s ) = a ϕ(m s )+s and a s 0 (d 0 ) s (d ) s ( ) s = a s ϕ( therefore m s )+ a s a s (mod m), for all a,m from Z (m 0) Observations: () If (a,m) = then d = Thus s = 0, and according to our theorem one has ϕ( m 0 )+ 0 a a 0 (mod m) therefore a ϕ( m 0 )+0 (mod m) But m = m 0 = m 0 = m 0 Thus: a ϕ( m) (mod m), and one obtains Euler s theorem () Let us have a and m two integers, m 0 and ( am, ) = d0, and m= m0d0 If ( ) =, then a ϕ( m 0 )+ a(mod m) Which, in fact, it results from our theorem with s = and m = m 0 This relation has a similar form to Fermat s theorem: a ϕ( p)+ a(mod p) C AN ALGORITHM TO SOLVE CONGRUENCIES One will construct an algorithm and will show the logic diagram allowing to calculate s and m s of the theorem Given as input: two integers a and m, m 0 It results as output: s and m s such that a ϕ( m s ) + s a s (mod m) Method: () A := a M := m i := 0 () Calculate d = (A, M ) and M ' = M / d (3) If d = take S = i and m s = M ' stop If d take A := d, M = M ' i := i +, and go to () 4
Remark: the accuracy of the algorithm results from lemma 3 end from the theorem See the flow chart on the following page In this flow chart, the SUBROUTINE LCD calculates D = (A, M ) and chooses D > 0 Application: In the resolution of the exercises one uses the theorem and the algorithm to calculate s and m s Example: 6 5604?(mod05765) One cannot apply Fermat or Euler because (6,05765)=3 One thus applies the algorithm to calculate s and m s and then the previous theorem: = (6,05765) = 3 m 0 = 05765 / 3 = 3555 i = 0; 3 thus i = 0 + =, d = (3,3555) =, m = 3555 / = 3555 Therefore 6 ϕ( 3555)+ 6 (mod05765) thus 6 5604 6 4 (mod05765) 5
Flow chart: START READ A, M A := A M := M I := 0 SUBROUTINE LCD (A,M,D) YES S = I NO MS = M D= WRITE S, MS I := I+ STOP M := M/D A := D 6
BIBLIOGRAPHY: [] Popovici, Constantin P Teoria numerelor, University Course, Bucharest, Editura didactică şi pedagogică, 973 [] Popovici, Constantin P Logica şi teoria numerelor, Editura didactică şi pedagogică, Bucharest, 970 [3] Creangă I, Cazacu C, Mihuţ P, Opaiţ Gh, Reischer Corina Introducere în teoria numerelor, Editura didactică şi pedagogică, Bucharest, 965 [4] Rusu E, - Aritmetica şi teoria numerelor, Editura didactică şi pedagogică, Editia a -a, Bucharest, 963 [Published in Bulet Univ Braşov, Series C, Vol XXIII, 98, pp 7-; MR: 84j:0006] 7