On the prime decomposition of integers of the form ( z n − y n )/(z−y)
aa r X i v : . [ m a t h . G M ] J un On the prime decomposition of integers of the form z n − y n z − y Rachid MarsliPreparatory Math DepartmentKing Fahd University of Petroleum and MineralsDhahran, 31261Kingdom of Saudi [email protected] 6, 2019
Abstract
In this work, the author shows a sufficient and necessary condition for aninteger of the form z n − y n z − y to be divisible by some perfect mth power p m ,where p is an odd prime and m is a positive integer. A constructive methodof this type of integers is explained with details and examples. Links be-tween the main result and known ideas such as Fermat’s last theorem, Goor-maghtigh conjecture and Mersenne numbers are discussed. Other relatedideas, examples and applications are provided. AMS Subj. Class.:11A07 ; 11D41Keywords: primitive root modulo integer; prime integer, perfect nth power; Fer-mat’s last theorem, Goormaghtigh conjucture.
Contrary to our expectations, while we were trying to prove Fermat’s last theoremby showing that if y and z are relatively prime and n and p are odd prime integers,then p n does not divide z n − y n z − y , we found that for almost every p we can constructinfinitely many integers of the form z n − y n z − y each of which is divisible by p n .1ot only that, but no matter how large is the positive integer m , we can alwaysconstruct integers of the form z n − y n z − y that are divisible by p m . The main tool ofour analysis in this work, is the concept of primitive root modulo integer. Givena positive integer p , we say that r is a primitive root modulo p if r is an integerrelatively prime to p and the smallest integer a such that r a ≡ mod p ) is φ ( p ) ,where φ denotes the well-known Euler function. A positive integer possesses aprimitive root if and only if n = 2 , , p t or p t , where p is an odd prime and t is apositive integer [4, Theorem 8.14]. Another important fact about primitive roots isgiven by the following theorem which we state as a lemma for its use in the proofof the main result. Lemma 1.1. [4, Theorem 8.9] Let p be an odd prime, then p k has a primitive rootfor all positive integer k . Moreover, if r is a primitive root modulo p , then r is aprimitive root modulo p k , for all positive integer k . Note that there are some rare cases where a primitive root modulo p is not aprimitive root modulo p . As an example, the prime integer p = 487 has a primitiveroot r = 10 which is not a primitive root modulo [4, Section 8.3]. Moreelementary ideas about primitive roots modulo integers can be found in numbertheory textbooks such as [1], [3], [4], [5], [6] and [9]. Throughout the paper, thegreatest common divisor of two integers a and b is denoted ( a, b ) . The following lemma is needed in the proof of the main result and containssome ideas that are well-known to mathematicians working on Fermat’s last theo-rem. Nevertheless, we prefer to provide a proof because we couldn’t find a refer-ence where all the three assertions of the lemma are proved together.
Lemma 2.1.
Let y and z be two relatively prime integers with z = y and let n bean odd prime integer.1. If n divides z − y , then (cid:16) z − y , z n − y n z − y (cid:17) = n .2. If n does not divides z − y , then n, ( z − y ) and z n − y n z − y are pairwiserelatively prime.3. n does not divide z n − y n z − y .Proof. We have z n = ( z − y + y ) n = n X i =2 (cid:18) ni (cid:19) ( z − y ) i y ( n − i ) + n ( z − y ) y ( n − + y n , z n − y n = ( z − y ) h n X i =2 (cid:18) ni (cid:19) ( z − y ) ( i − y ( n − i ) + ny ( n − i = ( z − y ) h ( z − y ) n n X i =2 (cid:18) ni (cid:19) ( z − y ) ( i − y ( n − i ) o + ny ( n − i , so that z n − y n z − y = ( z − y ) n n X i =2 (cid:18) ni (cid:19) ( z − y ) ( i − y ( n − i ) o + ny ( n − . (1)Since y and z are relatively prime, the power y n − and ( z − y ) are relatively prime.Hence, Formula (1) implies that (cid:16) z − y , z n − y n z − y (cid:17) = n, if n divides z − y, and (cid:16) z − y , z n − y n z − y (cid:17) = 1 , if n is relatively prime to z − y. Moreover, (1) can be rewritten as z n − y n z − y = ( z − y ) n − + n n − X i =1 (cid:18) ni (cid:19) ( z − y ) ( i − y ( n − i ) o . (2)Since n is a prime integer, we have (cid:16) n, (cid:18) ni (cid:19)(cid:17) = n for i = 1 , , . . . n − . (3)From (2) and (3) , we get z n − y n z − y ≡ ( z − y ) n − ( mod n ) . (4)It follows from (4) that if n is relatively prime to z − y , then n and z n − y n z − y arerelatively prime. This is to prove the second assertion. The third assertion of thelemma follows directly from the second one if n does not divide z − y . Otherwise,3uppose that n divides z − y . Then from (1) and (3), we can see easily that, in thiscase, z n − y n z − y ≡ ny ( n − ( mod n ) . (5)If n divides z n − y n z − y , then (5) implies that n divides y , so that also, n divides z since it divides z − y . This is in contradiction with our assumptions that y and z are relatively prime. Remark 2.2.
The first two assertions of Lemma 2.1 apply to the case where n = 2 ,but the third one does not. For example, if we take z = 5 , y = 3 and n = 2 , then divides − − .Next we state and prove the main result. Theorem 2.3.
Let y and z be two distinct nonnegative integers and let n be an oddprime integer. Let p be an odd prime integer that is different than n and relativelyprime to y . Let r be a primitive root modulo p and let m be a positive integer.Then p m divides z n − y n z − y if and only if n divides p − and z ≡ y r cp m − ( mod p m ) , where c is any integer that satisfies:1. < c < p − .2. p − divides nc .Proof. First recall that, by Lemma 1.1, r is also a primitive root modulo p m for m = 1 as well as for m = 3 , , . . . Suppose that n divides p − and z ≡ y r cp m − ( mod p m ) , (6)for some integer c such that < c < p − and p − divides nc . Formula (6)implies that z n ≡ y n r ncp m − ( mod p m ) . Since p − divides nc , it follows that φ ( p m ) , which is equal to ( p − p m − , divides ncp m − and therefore z n ≡ y n ( mod p m ) . (7)Also, Formula (6) implies that z ≡ y r cp m − ( mod p ) , which is equivalent to z ≡ y r c r c ( p m − − ( mod p ) . Since φ ( p ) , which is equal to p − , divides4 ( p m − − , it follows that z ≡ y r c ( mod p ) . By Lemma 1.1, r is a primitiveroot modulo p and since < c < p − , we have that r c mod p ) . Hence z y ( mod p ) . (8)It follows from (7) and (8) that z n − y n z − y is divisible by p m .Conversely, we have two different cases.1. Case1: z and y are relatively prime.Suppose that p m divides z n − y n z − y . Then p m divides z n − y n or equivalently, z n ≡ y n ( mod p m ) . (9)Since p is different than n and divides z n − y n z − y , Lemma 2.1 implies that p does not divides z − y . Hence, there exists an integer k such that < k < ( p − p m − (10)and z ≡ y r k ( mod p m ) . (11)This implies z n ≡ y n r nk ( mod p m ) . (12)From (9) and (12) we have y n (1 − r nk ) ≡ mod p m ) , which leads to (1 − r nk ) ≡ mod p m ) since y and p are relatively prime. Therefore, φ ( p m ) , which is equal to ( p − p m − , divides nk. (13)Since p = n , the above expression implies that p m − divides k and because < k < ( p − p m − , there exists an integer c such that < c < p − and k = cp m − . (14)From (14) and (13), we have that ( p − p m − divides ncp m − . Thus, ( p − divides nc. (15)Since < c < p − and n is a prime integer, Formula (15) implies that n divides p − , (16)We complete the proof of this case by taking (14) into (11) to obtain z ≡ y r cp m − ( mod p m ) . (17)5. Case2: ( z, y ) = q > .Let y ′ and z ′ be such that y = qy ′ and z = qz ′ . Then ( z ′ , y ′ ) = 1 and z n − y n z − y = q n − z ′ n − y ′ n z ′ − y ′ . (18)If p m divides z n − y n z − y with p and y being relatively prime, then p m divides z ′ n − y ′ n z ′ − y ′ . It follows, by Case1, that n divides p − and z ′ ≡ y ′ r cp m − ( mod p m ) ,so that z ≡ yr cp m − ( mod p m ) , where c is an integer such that < c
The integer c is even and different than p − . If c satisfies n c = p − , then the integer c takes all the values c , c = 2 c , c = 3 c , . . . , c n − =( n − c . That makes a total of ( n − values. Notice also that if z = yr c i P m − for some index i ∈ { , , . . . , n − } , then, by analogy between z and y , we have y = zr c j P m − for some integer j ∈ { , , . . . , n − } such that c i + c j = p − .Moreover, c i = c j for if they were equal, then we would have c = p − , which isimpossible as is already mentioned. Remark 2.5.
Note that, in the statement of Theorem 2.3, the condition p = n needs to be stated for the case m = 1 only. If m ≥ and p m divides z n − y n z − y ,then the third assertion of Lemma 2.1 ensures that p = n . Remark 2.6.
Observe that n divides p − in Theorem 2.3. Therefore, if p < n + 1 , then p m does not divide z n − y n z − y . Remark 2.7.
If an odd prime q divides z n − y n z − y but n does not divide q − , thenby Theorem 2.3 and Lemma 2.1, t is equal to n and divides z − y . Example 2.8.
Goormaghtigh conjecture states that the Diophantine equation x n − x − y n − y − , x > y > and n, m > , is satisfied for only two trivial cases: − − − − − − − − . The condition imposed by Theorem 2.3 that n divides p − is satisfied in bothcases. In the first case, we have n = 3 divides p − . In thesecond case, p = 8191 is a prime number and each of n = 3 and n = 13 divides p − )(7)(13) . Example 2.9.
A Mersenne number is an integer of the form n − .Therefore, it isof the form z n − y n z − y . It is well-known that if a prime p divides n − , where n isan odd prime, then n divides p − . This fact is in accordance with Theorem 2.3. Itmeans that for every odd prime integer n , there is another prime integer p strictlylarger than n . As it is known, This idea implies the infinitude of prime integers.Two particular cases of Theorem 2.3 are m = 1 and m = n . we state thesecond one as a corollary because of its connection with Fermat’s last theorem. Corollary 2.10.
Let y and z be two distinct nonnegative integers. Let p be an oddprime integer relatively prime to y and let r be a primitive root modulo p . and let n be an odd prime integer. Then p n divides z n − y n z − y if and only if n divides p − and z ≡ y r cp n − ( mod p n ) , where c is any integer that satisfies:1. < c < p − .2. p − divides nc . Corollary 2.11.
Let y and z be two distinct nonnegative integers and let n be anodd prime integer. Let p be an odd prime integer different than n , relatively primeto y and having the form p = 2 k + 1 for some positive integer k . Then p does notdivide z n − y n z − y . Proof.
Follows, immediately, from Theorem 2.3 since there is no odd prime integer n that divides p − k .As a completion of Theorem 2.3, we show that integers of the form z n − y n z − y are not divisible by , given that z and y are not both even and n is an odd primeinteger. 7 heorem 2.12. Let y and z be two distinct nonnegative integers not both even andlet n be an odd prime integer. Then does not divide z n − y n z − y .Proof. It suffices to show that z n − y n z − y is an odd integer. If one of y and z is oddand the other is even, then both ( z n − y n ) and ( z − y ) are odd integers. Hence, theirquotient z n − y n z − y is also odd. If each of y and z is odd, then ( z − y ) is even. Hence, z n − y n z − y has to be an odd integer since, by Lemma 2.1, (cid:16) z n − y n z − y , z − y (cid:17) = 1 or n . z n − y n z − y and divisible by p m Theorem 2.3, beside being a characteristic theorem, it is also a constructive the-orem. In other words, if y, p, n, m, are as in theorem 2.3, c = p − n and r , isa primitive root modulo p , then we can construct the set ξ ( y, p, n, m, c ) of allintegers of the form z n − y n z − y that are divisible by p m , ξ ( y, p, n, m, c ) = n z n − y n z − y (cid:12)(cid:12)(cid:12) z ≡ r c p m − ( mod p m ) o . (19)As we have explained in Remark 2.4, the integer c can be replaced by c i = i c for i = 1 , , . . . , n − , so that we can construct sets of the form: ξ ( y, p, n, m, c i ) = n z n − y n z − y (cid:12)(cid:12)(cid:12) z ≡ r i c p m − ( mod p m ) o , i = 1 , , . . . , n − . (20)The union, over i , of the above sets is ξ ( y, p, n, m ) = n − [ i =1 ξ ( y, p, n, m, c i ) . (21)Let ξ ( p, n, m ) be the set of all integers of the form z n − y n z − y that are divisible by p m , y relatively prime to p , Then ξ ( p, n, m ) is obtained by taking the union of the8ets of the form ξ ( y, p, n, m ) over all possible values of y . ξ ( p, n, m ) = [ y ∈ N p ∤ y ξ ( y, p, n, m ) = [ y ∈ N p ∤ y n − [ i =1 ξ ( y, p, n, m, c i ) . (22) Remark 3.1.
Unless p = 3 , there are many primitive roots that are incongruentmodulo p . However, we do not consider r to be a parameter in the construction of ξ ( p, n, m ) since this set remains invariant if we replace r by another primitive rootmodulo p . This can be easily verified.Suppose that z n − y n z − y ∈ ξ ( y, p, n, m, c = c ) . Then z ≡ y r c p m − ( mod p m ) . Since c = 2 c , the above congruence equation can be rewritten as z ≡ (cid:0) y r c p m − (cid:1) r c p m − ( mod p m ) . Letting y ′ = y r c p m − , we obtain z ≡ y ′ r c p m − ( mod p m ) , so that z n − y ′ n z − y ′ ∈ ξ ( y ′ , p, n, m, c ) . The above reasoning shows that [ y ∈ N p ∤ y n − [ i =1 ξ ( y, p, n, m, c i ) = [ y ∈ N p ∤ y ξ ( y, p, n, m, c ) . (23)Therefore, we have the following corollary. Corollary 3.2.
Let p be an odd prime integer for which there exists an other oddprime integer n such that p − n c for some positive integer c . Let r be aprimitive root modulo p . Then ξ ( p, n, m ) = [ y ∈ N p ∤ y n z n − y n z − y (cid:12)(cid:12)(cid:12) z ≡ r c p m − ( mod p m ) o (24)is the set of all integers of the form z n − y n z − y that are divisible by p m , where p doesnot divide y and m is a positive integer. 9 emark 3.3. Notice that no matter how the integer m is large, we can constructinfinitely many integers of the form z n − y n z − y divisible by p m . Notice also that ξ ( p, n, m ) ⊆ ξ ( p, n, m ′ ) , for ≤ m ′ < m. Example 3.4.
Let’s construct an integer of the form z − y z − y that is divisible by . Take p = 7 , r = 3 , n = 3 , c = 2 and y = 1 . We have that n = 3 divides p − and nc = 6 = p − . Construct the integer z = r c p n − = 3 . Then, byTheorem 2.3, = 343 divides (3 ) − − . Of course, this is a huge number. But Theorem 2.3 ensures that we can use positivenumbers that are less than and equivalent to z modulo p n . By the use of a calculator,we find easily that ≡
324 ( mod ) . Indeed, − − ) . Now, let’s ask a question:Is it true that, for an odd prime integer n , there are infinitely many odd primeintegers p such that n divides p − ?Consider the set ξ ( y = 1 , p, n, m = 1) = n z n − z − | z ≡ r c ( mod p ) o . and let E be the set of all odd prime integers p such that p divides some elementfrom ξ ( y = 1 , p, n, m = 1) . Since, by Theorem 2.3, n divides p − for everyelement p ∈ E , an affirmative answer of the above question can be obtained if weprove that there are infinitely many element in E . This seems to be true becauseevery two element of ξ ( y = 1 , p, n, m = 1) have, more likely, different primedecomposition. p m Beside its constructive aspect, Theorem 2.3 has other applications such as the fol-lowing.
Corollary 3.5.
Let p be an odd prime integer for which there exist another primeinteger n such that n divides p − . Let r be a primitive root modulo p . Let c be10n integer such that < c < p − and p − divides nc . Then, for every positiveinteger m , we have n − X k =0 r kcp m − ≡ mod p m ) . (25)In particular, for m = 1 , we have n − X k =0 r kc ≡ mod p ) . (26) Proof.
We choose an integer y relatively prime to p , and we construct the integer z = yr cp m − . (27)By Theorem 2.3, we have z n − y n z − y ≡ mod p m ) , which is equivalent to n − X k =0 z k y n − k − ≡ mod p m ) . (28)Taking (27) into (28), we obtain y n − n − X k =0 r kcp m − ≡ mod p m ) . (29)Since y and p are relatively prime, it follows from (29) that n − X k =0 r kcp m − ≡ mod p m ) . (30) Remark 3.6.
It is well-known that if r is primitive root mod p , then p − X k =0 r k ≡ mod p ) . (31)To see this, recall that r , r , . . . , ..., r n − form a complete residue set modulo p .A question that arises is: do we have similar formula for an integer t that is not aprimitive root modulo p ? The above corollary gives a partial answer to this questionby the mean of Formula (26) which can be considered as an extension of Formula1131). In fact, if t = r c , then t is not a primitive root modulo p since < c < p − and (cid:0) c, p − (cid:1) = 1 . Then Formula (26) becomes n − X k =0 t k ≡ mod p ) . (32)Note that n < p . That is, the number of summands in (32) is less than half of thatin (31). Example 3.7.
As in Example 3.4, we take p = 7 , r = 3 , n = 3 and c = 2 . If welet m = n = 3 , then we have P n − k =0 r kcp n − = P k =0 k = 1 + 3 + 3 ≡ ( mod ) ≡ − ( mod ≡ mod ≡ mod ) By the same reasoning, if m = 1 , then X k =0 kc = 3 + 3 + 3 = 91 ≡ mod . An other primitive root of is the integer . For m = 1 , we have X k =0 kc = 5 + 5 + 5 = 651 ≡ mod . mth power of a composite integer divides z n − y n z − y Let p , p , . . . , p k be k distinct prime integers each of which is different than n andrelatively prime to y . Suppose that the product k Y i =1 p m i i divides z n − y n z − y , where m , m , . . . , m k are k positive integers. According to Theorem 2.3, this hold if12nd only if n divides p i − for i = 1 , , . . . , k and z ≡ y r c p m − ( mod p m ) z ≡ y r c p m − ( mod p m ) . . .z ≡ y r c k p mk − k k ( mod p m k k ) , where, for i = 1 , , . . . , k , r i is a primitive root modulo p i , the integer c i satisfies < c i < p i − and p i − divides nc i . By the Chinese remainder theorem, theabove system of congruence equations holds if and only if z ≡ k X i =1 (cid:16) y r c i p mi − i i (cid:17) (cid:16) M i q i (cid:17) ( mod p m p m . . . p m k k ) ≡ y k X i =1 M i q i r c i p mi − i i ( mod p m p m . . . p m k k ) , (33)where M i = Q kj =1 p m j j p m i i and q i is any integer that satisfies M i q i ≡ mod p m i i ) . The following corollary summarize the above result.
Corollary 3.8.
Let y and z be two relatively prime integers and let n be an oddprime integer. Let p , p , . . . , p k be k distinct odd prime integers, each of whichis different than n . and let r , r , . . . , r k be, respectively, primitive root mod-ulo p , p , . . . , p k . Let m , m , . . . , m k be k positive integers. Then the product k Y i =1 p m i i divides z n − y n z − y if and only if n divides p i − , for i = 1 , , . . . , k, (34)and z ≡ y k X i =1 M i q i r c i p mi − i i ( mod p m p m . . . p m k k ) , (35)where1. M i = Q kj =1 p m j j p m i i , q i is any integer that satisfies M i q i ≡ mod p m i i ) , c i is an integer such that < c i < p i − and p i − divides nc i . Fermat’s last theorem [8] states:
Theorem 4.1.
For every positive integer n with n ≥ , no positive integers x, y and z satisfy z n = x n + y n . This theorem, which has been proved around 1995 [8], implies the followingfact.
Corollary 4.2.
Let z and y be two relatively prime integers, and let n be an oddprime integer. Then z − y is a perfect nth power if and only if z n − y n z − y is not aperfect nth power. In particular, if z − y = 1 , then z n − y n z − y is not an nth perfectpower.We have proved in this work, that z n − y n z − y can be multiple of some perfect nth power p n . But we don’t know if z n − y n z − y , itself, can be a perfect nth power havingthe form ( p p . . . ) n for not necessary distinct prime integers p , p , . . . By goingback to Formula (24) and looking at how large is the set ξ ( p, n, m ) and the degreeof freedom that we have to construct such set by acting on different parameters p and n , one may believe that there is chance for some elements of ξ ( p, n, m ) to beperfect nth powers. For instance, consider the number a = 16 − − which isequal to . Of course, the integer a is not a perfect nth power since n = 3 . Butit is well a perfect power. Moreover, it is a perfect power of a prime integer. Sincesuch number a exists and it is remarkably small, we believe that nothing impedethe existence of a perfect nth power of the form z n − y n z − y . However, it may turnout that the smallest of these numbers is tremendously large and therefore difficultto reach with a computer. Perhaps, a constructive proof, is the best way to find suchintegers if they exist.For mathematicians seeking a proof of Fermat’s last theorem by the mean of clas-sical methods, we have a little result that may be of some use and which is conse-quence of Theorem 2.3. 14 heorem 4.3. Suppose that there are pairwise relatively prime positive integers x, y, z such that z n = x n + y n , where n is an odd prime integer. If p is an oddprime integer such that p = n and p divides x n y n z n ( z − x )( z − y )( x + y ) , then n divides p − .Proof. p divides one of x n z − x , y n z − y and z n x + y . Then, by Theorem 2.3, n divides p − . We believe that a lot can be done about the integers of the form z n − y n z − y . A betterunderstanding of this type of integers may lead to a more accessible proof of Fer-mat’s last theorem as well as to the solutions of other Diophantine equations. Forinstance, an observations made on some few integers of the form z n − y n z − y gives usthe impression that there may be always some prime integer p divisor of z n − y n z − y that is greater than each of | z | and | y | . We strongly believe that this observationholds for all integers of the form z n − y n z − y . Therefore we state it as a conjecture. Conjecture 5.1.
Let z and y be two relatively prime positive integers with z > y and let n be an odd prime integer. There is a prime integer p divisor of z n − y n z − y such that p > z . If it happens that this conjecture is true, then Fermat’s last theorem will be animmediate consequence of it. 15 z z − y z − y prime decomp p, p > z y z z − y z − y prime decomp p, p > z ∗
13 13 7 8 15961 11 ∗ prime
109 7 9 21121 prime ∗
43 43 7 10 27731 11 ∗ prime
151 7 11 36061 prime ∗
67 67 7 12 46405 5 ∗ prime
229 7 13 59081 11 ∗ ∗
131 131 ,
415 13 259 7 ∗
37 37 7 15 92821 prime ∗
97 97 7 16 114641 prime ∗
19 19 7 17 140305 5 ∗ ∗ ∗ ∗
19 19 7 18 170251 61 ∗ prime
439 7 19 204941 11 ∗ ∗
601 601 ,
315 19 481 13 ∗
37 37 7 20 244861 prime prime
571 7 22 342455 5 ∗ prime
619 7 23 401221 71 ∗ , ∗
223 223 7 24 467401 11 x ∗
103 103 7 25 541601 31 x , ∗
277 277 7 26 624451 prime ∗
127 127 7 27 716605 5 ∗ ∗
571 251 , ∗
73 73 7 29 931561 41 ∗ , ∗
337 337 7 30 1055791 11 ∗ ∗ , ∗
163 163 7 31 1192181 prime
Table1: The prime decomposition of some small numbers of the forms z n − y n z − y .Each one of them has a prime divisor that is larger than z . Acknowledgements
The author would like to thank Abdullah Laaradji from King Fahd Universityof Minerals and Petroleum for his very useful comments and suggestions.