A Study on Erdős-Straus conjecture on Diophantine equation \frac{4}{n}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}
aa r X i v : . [ m a t h . G M ] D ec A Study on Erd˝os-Straus conjecture onDiophantine equation n = x + y + z ∗ S. Maiti , † Department of Mathematics, The LNM Institute of Information TechnologyJaipur 302031, India Department of Mathematical Sciences, Indian Institute of Technology (BHU),
Varanasi-221005, India
Abstract
The Erd˝os-Straus conjecture is a renowned problem which describes that for everynatural number n ( ≥ n can be represented as the sum of three unit fractions. Themain purpose of this study is to show that the Erd˝os-Straus conjecture is true. Thestudy also re-demonstrates Mordell theorem which states that n has a expression as thesum of three unit fractions for every number n except possibly for those primes of theform n ≡ r (mod 780) with r = 1 , , , , , . For l, r, a ∈ N ; l +1 − l + r = r − l + r )(24 l +1) with 1 ≤ r ≤ l , if at least one of the sums in right side of the expression,say, a + (4 r − a − , ≤ a ≤ r − r such that a, (4 r − a −
1) divide (6 l + r )(24 l + 1); then the conjecture is valid for the corresponding l .However, in this way the conjecture can not be proved only twelve values of l for l up to l = 10 . Keywords: Erd˝os-Straus Conjecture, Diophantine Equation; n = x + y + z ; ElementaryNumber Theory. The famous Erd˝os-Straus conjecture in number theory, formulated by Paul Erd˝os and Ernst G.Strauss [1, 2] in 1948, states that for every natural number n ( ≥ ∃ natural numbers x, y, z ∗ AMS 2010 Mathematics Subject Classification: 11Dxx, 11D68, 11N37 (Primary) 11D45, 11Gxx, 14Gxx;11D72, 11N56, 11P81 (Secondary) † Corresponding author, Email address: [email protected]/[email protected] (S. Maiti) n can be expressed as 4 n = 1 x + 1 y + 1 z . (1)Many researchers, not only in Number Theory but also in different areas of Mathematics, gaveattention to this conjecture such as L. Bernstein [3], M. B. Crawford [4], M. Di Giovanni, S.Gallipoli, M. Gionfriddo [5], C. Elsholtz and T. Tao [6], J. Ghanouchi [7, 8], L. J. Mordell [9],D. J. Negash [10], R. Obl´ath [11], L. A. Rosati [12], J. W. Sander [13], R. C. Vaughan [14], K.Yamamoto [15], J. W. Porras Ferreira [19, 20] etc. The validity of the conjecture for all n ≤ and n ≤ was reported by Swett [16] and Salez [17] respectively.If m and n are relatively prime integers, then Schinzel [18] established that mt + n = x ( t ) + y ( t ) + z ( t ) having x ( t ), y ( t ) and z ( t ) as integer polynomials in t together with positive leadingcoefficients and non quadratic residue n (mod m ). Mordell [9] demonstrated that the validityof the Erd˝os-Straus conjecture for all n except possible cases where n is congruent to 1 , 11 ,13 , 17 , 19 or 23 (mod 840).The conjecture can be proved if it is derived for the all prime numbers to any of the followingcases (i) n = p = 2 m + 1 (ii) n = p = 4 m + 1 (iii) n = p = 8 m + 1 (iv) n = p = 24 m + 1 where m ∈ N . For l, r, a ∈ N ; l +1 − l + r = r − l + r )(24 l +1) with 1 ≤ r ≤ l . If at least one of the sums inright side of the expression, say, a + (4 r − a − , ≤ a ≤ r − r such that a, (4 r − a −
1) divide (6 l + r )(24 l + 1); then l +1 has the expression in theform of the equation (1). However, in this way the expression in the form of equation (1) cannot be proved for all values of l although it can be established for most of the values of l . Thecomputations of l +1 − l + r = r − l + r )(24 l +1) for l up to l = 10 has been carried out which showsthat the expression in the form of the equation (1) with n = 24 l + 1 can not be proved onlytwelve values of l . Finally, it has been shown (by other way) in Section 5 that the conjecture istrue. If we want to express any n ∈ N in the form of (1), then it is equivalent to find the equation(1) for all primes n = p . We know that (i) if n = 2 m , then n = m = m + m + m (ii)if n = 3 m , then n = m = m + m + m = m + m +1) + m ( m +1) = m + m +1 + m ( m +1) (iii) if n = 3 m + 2, then n = m +2 = m +2 + m +1 + m +1)(3 m +2) (iv) if n = 4 m + 3, then2 n = m +3 = m +1 + m +3)( m +1) + m +3)( m +1) (v) if n = 4 m and n = 4 m + 2, then these reduceto case (i) where m ∈ N . Thus we have to prove the equation (1) for all n = 4 m + 1.From the equation (1), we get n = x + y + z > x , y , z i.e. x , y , z ≤ n ]+1 as n − n ] ≤
0. If x ≤ y ≤ z , then n ≤ x i.e. n ] ≤ x . Thus n ] ≤ x ≤ n ]+1 and z ≤ y ≤ x .For n − x = x − nnx with x ≤ y ≤ z and n ] ≤ x ≤ n ]+1 , if 4 x − n = p α i p α i · · · p α r i r + p β j p β j · · · p β s j s with nx = p γ p γ · · · p γ m m , { p i , p i , · · · , p i r ; p j , p j , · · · , p j s } ⊂ { p , p , · · · , p m } ; p α i p α i · · · p α r i r , p β j p β j · · · p β s j s divide nx ; then n has a expression as in the form of (1).Again if 4 x − n = y + z ; y , z ∈ N with ( y , z ) = 1, then n − x = x − nnx = y + z nx = nxy + nxz . If n satisfies the equation (1), then nx = k y , k ∈ N and nx = k z , k ∈ N i.e. k y = k z . Or, k = gy , g ∈ N as ( y , z ) = 1. Then k = gz and nx = gy z . Thus n = x + gz + gy = x + y + z with g = nxy z , y = gy and z = gz .If 4 x − n = dy + dz ; d, y , z ∈ N with ( y , z ) = 1, then n − x = x − nnx = dy + dz nx = nxdy + nxdz .If n satisfies the equation (1), then nx = dk y , k ∈ N and nx = dk z , k ∈ N i.e. k y = k z .Or, k = gy , g ∈ N . Then k = gz and nx = gdy z . Thus n = x + gz + gy = x + y + z with g = nxdy z , y = gy and z = gz .For m +1 , [ m +14 ] = m and [ m +1)4 ] = 3 m , m ≤ x ≤ m +1 . Then for the equation (1) with n = 4 m + 1, the possible cases of x will be only x = m +1 , m +2 , · · · , m . Thus for44 m + 1 − m + r = 4 r − m + r )(4 m + 1) = { r − } , { r − } , · · · , { (2 r −
1) + 2 r } ( m + r )(4 m + 1) (2)with r ∈ N and 1 ≤ r ≤ m ; if at least one of the sums in right side of the equation, say, a + (4 r − a − , ≤ a ≤ r − r such that a, (4 r − a − m + r )(4 m + 1); then m +1 satisfies the equation (1). A question is arising naturallythat whether the equation (1) can be proved with the help of the equation (2)? The answer ofthis question will be given later.If m = 2 k −
1, then 4 m + 1 = 8 k −
3, [ k − ] = 2 k − , [ k − ] = 6 k − , k − ≤ x ≤ k .So, k − − k = k (8 k − = k (8 k − = k (8 k − + k (8 k − i.e. k − = k + k (8 k − + k (8 k − .If m = 2 k , then 4 m + 1 = 8 k + 1, [ k +14 ] = 2 k, [ k +1)4 ] = 6 k, k ≤ x ≤ k +1 . So, k +1 − k +1 = k +1)(8 k +1) .Again if k = 3 l −
2, then 8 k + 1 = 24 l −
15, [ l − ] = 6 l − , [ l − ] = 18 l − , l − ≤ x ≤ l − . So, l − − l − = l − r − = l − l − = l − l − + l − l − i.e. l − = l − + l − l − + l − l − .If k = 3 l −
1, then 8 k + 1 = 24 l −
7, [ l − ] = 6 l − , [ l − ] = 18 l − , l − ≤ x ≤ l − .3o, l − − l − = l − l − = l l (6 l − l − = l − l (6 l − l − = l (6 l − l − + l (24 l − i.e. l − = l − + l (6 l − l − + l (24 l − .If k = 3 l , then 8 k + 1 = 24 l + 1, [ l +14 ] = 6 l, [ l +1)4 ] = 18 l . Thus we have to prove theequation (1) for all n = 24 l + 1 with l ≤ x ≤ l +1 . Using the expression of the equation (2),we get424 l + 1 − l + r = 4 r − l + r )(24 l + 1) = { r − } , { r − } , · · · , { (2 r −
1) + 2 r } (6 l + r )(24 l + 1) (3)with r ∈ N and 1 ≤ r ≤ l . If at least one of the sums in right side of the expression, say, a + (4 r − a − , ≤ a ≤ r − r such that a, (4 r − a − l + r )(24 l + 1); then l +1 has the expression in the form of the equation (1). Thus thesame question is arising naturally that whether the equation (1) can be proved with the help ofthe equation (3)?The answer of this question is that the expression in the form of equation (1) can not beproved with the help of the equation (3) for all values of l ∈ N . However, the equation (1) canbe proved with the help of the equation (3) for most of the values of l ∈ N and it can not beproved for very less number of values of l ∈ N . To find the answer, the computations of l up to l = 10 has been carried out which shows that the expression in the form of the equation (1)with n = 24 l + 1 can not be proved with the help of the equation (3) only if n = (i.e. for l = 17), n = (i.e. for l = 24), n = (i.e. for l = 232), n = (i.e. for l = 400), n = (i.e. for l = 997), n = (i.e. for l = 3477), n = (i.e. for l = 4250), n = (i.e.for l = 13734), n = (i.e. for l = 29680), n = (i.e. for l = 47260), n = (i.e.for l = 71842), n = (i.e. for l = 71925) and the computations are taking several days togive result in Mathematica in my system with 8 GB RAM and 1 TB har disk if the equation(1) with n = 24 l + 1 can not be proved with the help of the equation (3) for l ≥ r ∈ N , on numerator, denominator ofright side of l +1 − l + r = r − l + r )(24 l +1) such that l +1 − l + r = r − l + r )(24 l +1) = (4 r − r r (6 l + r )(24 l +1) and atleast one of the sums of numerator (4 r − r in right side of the expression, say, a +((4 r − r − a )for at least one of the possible value of r such that a, (4 r − r − a divide r (6 l + r )(24 l + 1);then l +1 has the expression in the form of the equation (1). However, there is no way to find r , r in this process but trial and error method.A suitable method to find the expression of the equation (1) for all n ∈ N has been discussedin Section 5. 4 .1.1 Example (i) (for l = 17) − = × × = × × × = × × = + i.e. = + + .(ii) (for l = 24) − = × × = × × × × = × × × = + i.e. = + + .(iii) (for l = 232) − = × × × = × × × × × = × × × × = + i.e. = + + .(iv) (for l = 400) − = × × × = × × × × × = × × × × = + i.e. = + + .(v) (for l = 997) − = × × × = × × × × × = × × × × = + i.e. = + + .(vi) (for l = 3477) − = × × × × = × × × × × × = × × × × × = + i.e. = + + .(vii) (for l = 4250) − = × × × = × × × × × = × × × × = + i.e. = + + .(viii) (for l = 13734) − = × × = × × × × = × × × = + i.e. = + + .(ix) (for l = 29680) − = × × × × = × × × × × × = × × × × × = + i.e. = + + .(x) (for l = 47260) − = × × = × × × × = × × × = + i.e. = + + .(xi) (for l = 71842) − = × × × × = × × × × × × = × × × × × = + i.e. = + + .(xii) (for l = 71925) − = × × × × = × × × × × × = × × × × × = + i.e. = + + . If 6 l + 1 or 24 l + 1 has a factor 3 b + 2 ( l, b ∈ N ), then l +1 has the expression in the form of (1). Proof : If 6 l + 1 or 24 l + 1 ( l ∈ N ) has a factor 3 b + 2,then l +1 − l +1 = b +2) × f (where (24 l + 1)(6 l + 1) = f × (3 b + 2))= b +1)(3 b +2) × ( b +1) × f = b +2)(3 b +2) × ( b +1) × f = b +2) × ( b +1) × f + b +1) × f i.e. l +1 = l +1 + b +2) × ( b +1) × f + b +1) × f . 5xample: − = × ( b = 1) × × × = × × = + . If 3 l + 1 = 5 b ( l, b ∈ N ), then b − has the expression in the form of (1). Proof : If 3 l + 1 = 5 b ( l, b ∈ N ), then l +1 − l +2 = b − − b = b (40 b − = b (40 b − = b (40 b − + b (40 b − i.e. b − = b + b (40 b − + b (40 b − . Note:
If 24 l + 1 has a factor 5 ( l ∈ N ), then l +1 has the expression in the form of (1). If 3 l + 1 = 7 b ( l, b ∈ N ), then b − has the expression in the form of (1). Proof : If 3 l + 1 = 7 b ( l, b ∈ N ), then l +1 − l +2 = b − − b = b (56 b − = b (56 b − = b (56 b − + b (56 b − i.e. b − = b + b (56 b − + b (56 b − . Note:
If 24 l + 1 has a factor 7 ( l ∈ N ), then l +1 has the expression in the form of (1). If 3 l + 1 = 7 b + 5 ( l, b ∈ N ), then b +33 has the expression in the form of (1). Proof : If 3 l + 1 = 7 b + 5 ( l, b ∈ N ), then l +1 − l +2 = b +33 − b +5) = b +5)(56 b +33) = b +1)2( b +1)(7 b +5)(56 b +33) = b +5)2( b +1)(7 b +5)(56 b +33) = b +1)(7 b +5)(56 b +33) + b +1)(56 b +33) i.e. b +33 = b +5) + b +1)(7 b +5)(56 b +33) + b +1)(56 b +33) . Note:
If 24 l + 1 has a factor 7 b + 5 ( l, b ∈ N ), then l +1 has the expression in the form of(1). If 3 l + 1 = 7 b + 6 ( l, b ∈ N ), then b +41 has the expression in the form of (1). Proof : If 3 l + 1 = 7 b + 6 ( l, b ∈ N ), then l +1 − l +2 = b +41 − b +6) = b +6)(56 b +41) = b +1)2( b +1)(7 b +6)(56 b +41) = b +6)2( b +1)(7 b +6)(56 b +41) = b +1)(7 b +6)(56 b +41) + b +1)(56 b +41) i.e. b +41 = b +6) + b +1)(7 b +6)(56 b +41) + b +1)(56 b +41) . Note:
If 24 l + 1 has a factor 7 b + 6 ( l, b ∈ N ), then l +1 has the expression in the form of(1). 6 .7 Lemma If 3 l + 1 = 7 b + 3 ( l, b ∈ N ), then b +17 has the expression in the form of (1). Proof : If 3 l + 1 = 7 b + 3 ( l, b ∈ N ), then l +1 − l +2 = b +17 − b +3) = b +3)(56 b +17) = b +1)2(2 b +1)(7 b +3)(56 b +17) = b +6)(2 b +1)(14 b +6)(56 b +41) = b +1)(7 b +3)(56 b +17) + b +1)(56 b +17) i.e. b +17 = b +3) + b +1)(7 b +3)(56 b +17) + b +1)(56 b +17) . Note:
If 24 l + 1 has a factor 7 b + 3 ( l, b ∈ N ), then l +1 has the expression in the form of(1). The expression in the form of (1) has a solution for every natural number n , except possibly forthose primes of the form n ≡ r (mod 120), with r = 1 , . Proof : (i) If l = 5 b − l, b ∈ N ), then l +1 − l +1 = b − − b − = b − b − = × × b − b − = b − b − = b − b − + b − b − i.e. b − = b − + b − b − + b − b − .(ii) If l = 5 b − l, b ∈ N ), then l +1 − l +2 = b − − b − = b − b − = b − b − = b − b − + b − b − i.e. b − = b − + b − b − + b − b − .(iii) If l = 5 b − l, b ∈ N ), then l +1 − l +1 = b − − b − = b − b − = × × b − b − = b − b − = b − b − + b − b − i.e. b − = b − + b − b − + b − b − .(iv) If l = 5 b − l, b ∈ N ), then l +1 = b − where 120 b −
71 = 49 (mod 120)(v) If l = 5 b ( l, b ∈ N ), then l +1 = b +1 where 120 b + 1 = 1 (mod 120).Hence, the expression in the form of (1) has a solution for every natural number n , exceptpossibly for those primes of the form n ≡ r (mod 120), with r = 1 , . The expression in the form of (1) has a solution for every number n , except possibly for thoseprimes of the form n ≡ r (mod 780), with r = 1 , , , , , . Proof : From Theorem 3, we know that the expression (1) has a solution if n = b + r except r = 1 , − n = b +1 . 7i) If b = 7 c − b, c ∈ N ), then b +1 = c − where 840 c −
719 = 11 (mod 840).(ii) If b = 7 c − b, c ∈ N ), then b +1 − b +3 = c − − c − = c − c − = × × c − c − = c − c − = c − c − + c − c − i.e. c − = c − + c − c − + c − c − .(iii) If b = 7 c − b, c ∈ N ), then b +1 = c − where 840 c −
479 = 19 (mod 840).(iv) If b = 7 c − b, c ∈ N ), then b +1 − b +2 = c − − c − = c − c − = (15 c − × c − × c − c − = c − c − c − c − = c − c − c − + c − c − i.e. c − = c − + c − c − c − + c − c − .(v) If b = 7 c − b, c ∈ N ), then b +1 − b +2 = c − − c − = c − c − = (15 c − × c − × c − c − = c − c − c − c − = c − c − c − + c − c − i.e. c − = c − + c − c − c − + c − c − .(vi) If b = 7 c − b, c ∈ N ), then b +1 − b +2 = c − − c − = c − c − = c − c − = c − c − + c − c − i.e. c − = c − + c − c − + c − c − .(vii) If b = 7 c ( b, c ∈ N ), then b +1 = c +1 where 840 c + 1 = 1 (mod 840).(2) Let n = b − .(i) If b = 7 c − b, c ∈ N ), then b − − b − = c − − c − = c − c − = c − c − = c − c − + c − c − i.e. c − = c − + c − c − + c − c − .(ii) If b = 7 c − b, c ∈ N ), then b − = c − where 840 c −
671 = 13 (mod 840).(iii) If b = 7 c − b, c ∈ N ), then b − = c − where 840 c −
551 = 17 (mod 840).(iv) If b = 7 c − b, c ∈ N ), then b − − b − = c − − c − = c − c − = (30 c − × c − × c − c − = c − c − c − c − = c − c − c − + c − c − i.e. c − = c − + c − c − c − + c − c − .(v) If b = 7 c − b, c ∈ N ), then b − = c − where 840 c −
311 = 23 (mod 840).(vi) If b = 7 c − b, c ∈ N ), then b − − b − = c − − c − = c − c − = (15 c − × c − × c − c − = c − c − c − c − = c − c − c − + c − c − i.e. c − = c − + c − c − c − + c − c − .(vii) If b = 7 c ( b, c ∈ N ), then b − − b − = c − − c − = c − c − = (15 c − × c − × c − c − = c − c − c − c − = c − c − c − + c − c − i.e. c − = c − + c − c − c − + c − c − .Hence, the expression in the form of the equation (1) has a solution for every number n ,except possibly for those primes of the form n ≡ r (mod 840), with r = 1 , , , , , .8 Theorem
The Erd˝os-Straus conjecture is true i.e. the expression in the form of (1) has a solution for everynatural number n . Proof : We know that the Erd˝os-Straus conjecture is equivalent to find the equation (1)for all primes. Because, it is well known that if n = x + y + z is true for n = p , then we get mp = mx + my + mz for any natural number m . Let us consider, n as prime number p , which isgreater than 2, in the equation (1). For p = 2, we know = + + . Case 1 : If x = y = z , then 3 p = 4 x which is a invalid equation. Thus the equation (1) hasno solution if n is a prime number and x = y = z . Case 2 : Two of x, y, z are equal. Without loss of generality, let x = y .Then 4 p = 2 x + 1 z i.e. p ( x + 2 z ) = 4 xz. (4)Thus p divides x or p divides y for p > p divides x , then x = up, u ∈ N . From (4), we get up +2 z = 4 uz . Or, 2 z (2 u −
1) = up .(i) If p divides z i.e. z = vp, v ∈ N , then 2 v (2 u −
1) = u , which is a invalid equation.(ii) If p divides 2 u − u − v p, v ∈ N , then 2 zv = u and u is even since z = u v i.e. 2 v divides u . Then u = 2 u , 4 u − v p and v divides u . So, v divides u and v divides 4 u − v = 1. Thus p = 4 u − u + 3 , x = 2( u + 1) p = y, z = u + 1,where u = u + 1.Then 4 p = 12( u + 1) p + 12( u + 1) p + 1 u + 1 = 1 p ( p +1)2 + 1 p ( p +1)2 + 1 p +14 . (5)(II) Let p divides z , then z = u p, u ∈ N . From (4), we get x + 2 u p = 4 u x . Or,(4 u − x = 2 u p .(i) If p divides x i.e. x = v p, v ∈ N , then (4 u − v = 2 u , which is a invalid equation.(ii) If p divides 4 u − u − v p, v ∈ N , then xv = 2 u . Thus v divides 2 u and v divides 4 u − v = 1. Thus p = 4 u − u + 3 , x = 2( u + 1) = y, z = ( u + 1) p ,where u = u + 1.Then 4 p = 12( u + 1) + 12( u + 1) + 1( u + 1) p = 1 ( p +1)2 + 1 ( p +1)2 + 1 p ( p +1)4 . (6) Case 3 : Let x = y = z = x . From the equation (1), we get p ( xy + yz + zx ) = 4 xyz . Then p divides at least one of x, y, z . Without loss of generality, let p divides x i.e. x = u p, u ∈ N .Thus u p ( y + z ) = yz (4 u − . (7)9I) Let p divides y or z . Without loss of generality, let p divides y . Then y = v p, v ∈ N .From (7), we get u v p = z (4 u v − u − v ).(i) If p divides z i.e. z = w p, w ∈ N , then u v + w v + w u = 4 u v w , which is ainvalid equation.(ii) If p divides 4 u v − u − v ,then 4 u v − u − v = w p, w ∈ N . (8)So, u v = zw . Thus w divides u v , w divides 4 u v − u − v and hence w divide u + v , u , v , u ( u − v ), v ( u − v ), ( u − v ) etc. Now, w divide u v and u + v mean u v = w w , w ∈ N and u + v = w w , w ∈ N . Then from (8), we get p = 4 w − w . Hence theequation (1) has solution when n = p = 4 w − w with x = u p , y = v p , z = w , u v = w w , u + v = w w since1 x + 1 y + 1 z = ( u + v ) w + u v pu v w p = w w w + w w pu v w p = ( w + p ) w u v p = 4 w w w w p = 4 p . (9)We have to demonstrate the solutions of the equation (1) if n = 4 m + 1 since we already knowsolutions of it for n = 4 m , n = 4 m + 2 and n = 4 m + 3.(a) If w = 3, then p = 4 w − n = 4 m +1), v = 3 w − u , w w = u v = u (3 w − u )i.e. w = u u − w = u u − p +34 . Thus we have to choose u (cid:18) > (cid:20) ( p +34 ) (cid:21)(cid:19) such that w ∈ N .However, it will have limited use for the solutions when n = 4 m + 1 as w = 3 is a special case.For example, if p = 13, then w = 4 , w = 3. If we choose u = 2, then w = 2 , v = 4.Hence = × + × + . We can try to reproduce the solutions of Section 2.1.1.For p = 409, w = 103 , w = 3. Then, we can not find any w ∈ N if u < p = 577, w = 145 , w = 3. If we choose u = 50, then we can get w = 500 , v = 1450.Thus = × + × + . If we choose u = 58, then we can get w = 116 , v = 290and = × + × + .For p = 5569, w = 1393 , w = 3. Then, we can not find any w ∈ N if u < p = 9601, w = 2401 , w = 3. Then, we can not find any w ∈ N if u < Note : Let w ∈ N , then 3 u − w >
0. Thus v = 3 w − u = u u − w − u = u w u − w > w ∈ N implies v ∈ N .(b) Let w = 4 w +3 where w = 0 or w ∈ N , then p = 4( w − w ) − n = p = 4 m +1).So, w − w = p +34 ∈ N . Or, w = w + p +34 , w = u (4 w +3) u − p +34 − w . Thus, we have to choose w , u (cid:16) > h w + p +34 w +3 i(cid:17) such that w ∈ N . Using these w , u , w ; we can calculate w = w + p +34 ,10 = (4 w + 3) w − u . Hence, in this way we can generate all solutions of the equation (1) if n = p = 4 m + 1.As examples, we can reproduce the solutions of Section 2.1.1. We can get multiple solutionsif we choose w , u ≤ u > h w + p +34 w +3 i , however for same solutions as of Section 2.1.1,we have to increase the range of u for majority of cases.Let p = 409. For w , u ≤ w = 1 , u = 15. Then w = 225, v = 1560, w = 104 and hence = × + × + . The last solution is for w = 14 , u = 234. Then w = 4, v = 2, w = 117 and hence = × + × + . The same solutions as of Section 2.1.1 is for w = 1 , u = 16. Then w = 32, v = 208, w = 104 and hence = × + × + .Let p = 577. For w , u ≤ w = 0 , u = 50. Then w = 500, v = 1450, w = 145 and hence = × + × + . The last solution is for w = 20 , u = 330. Then w = 4, v = 2, w = 165 and hence = × + × + . The same solutions as of Section 2.1.1 is for w = 0 , u = 58. Then w = 116, v = 290, w = 104 and hence = × + × + .Let p = 5569. For w , u ≤ w = 1 , u = 204. Then w = 1224, v = 8364, w = 1394 and hence = × + × + , which is the same solutions as of Section 2.1.1. The lastsolution is for w = 35 , u = 10. Then w = 50, v = 7140, w = 1428 and hence = × + × + .Let p = 9601. For w , u ≤ w = 4 , u = 130. Then w = 260, v = 4810, w = 2405 and hence = × + × + , which is the same solutions as of Section 2.1.1. The last solution is for w = 104 , u = 6. Then w = 4, v = 1670, w = 2505 and hence = × + × + .Let p = 23929. For w , u ≤ w = 1 , u = 855. Then w = 731025, v = 5116320, w = 5984 and hence = × + × + . The last solution is for w = 854 , u = 2. Then w = 4, v = 13674, w = 6837 and hence = × + × + . To get the same solutionsas of Section 2.1.1, we can consider the case w ≤ , u ≤ w = 1 , u = 1056.Then w = 792, v = 4488, w = 5984 and hence = × + × + .Let p = 83449. For w , u ≤ w = 5 , u = 908. Then w = 51529, v = 1184259, w = 20868 and hence11 = × + × + . The last solution is for w = 353 , u = 15. Then w = 25, v = 35360, w = 21216 and hence = × + × + . To get the same solutionsas of Section 2.1.1, we can consider the case w ≤ , u ≤ w = 2 , u = 1950.Then w = 6500, v = 69550, w = 20865 and hence = × + × + .Let p = 102001. For w , u ≤ w = 7 , u = 826. Then w = 6962, v = 214996, w = 25508 and hence = × + × + . The last solution is for w = 542 , u = 12. Then w = 16, v = 34724, w = 26043 and hence = × + × + . Toget the same solutions as of Section 2.1.1, we can consider the case w ≤ , u ≤ w = 1 , u = 3732. Then w = 22392, v = 153012, w = 25502 and hence = × + × + .Let p = 329617. For w , u ≤ w = 26 , u = 774. Then w = 1548, v = 164862, w = 82431 and hence = × + × + . The last solution is for w = 867 , u = 24. Then w = 18, v = 62454, w = 83272 and hence = × + × + . To getthe same solutions as of Section 2.1.1, we can consider the case w ≤ , u ≤ w = 0 , u = 32962. Then w = 65924, v = 164810, w = 82405 and hence = × + × + .Let p = 712321. For w , u ≤ w = 47 , u = 936. Then w = 1352, v = 257296, w = 178128 and hence = × + × + . The last solution is for w = 587 , u = 76. Then w = 722, v = 1697346, w = 178668 and hence = × + × + . Toget the same solutions as of Section 2.1.1, we can consider the case w ≤ , u ≤ w = 5 , u = 7974. Then w = 11961, v = 267129, w = 178086 and hence = × + × + .Let p = 1134241. For w , u ≤ w = 101 , u = 697. Then w = 28577, v = 11630142, w = 283662 and hence = × + × + . The last solution is for w = 696 , u = 102. Then w = 612, v = 1705542, w = 284257 and hence = × + × + .12o get the same solutions as of Section 2.1.1, we can consider the case w ≤ , u ≤ w = 0 , u = 94926. Then w = 7404228, v = 22117758, w = 283561 and hence = × + × + .Let p = 1724209. For w , u ≤ w = 125 , u = 858. Then w = 1859, v = 934219, w = 431178 and hence = × + × + . The last solution is for w = 749 , u = 144. Then w = 384, v = 1151472, w = 431802 and hence = × + × + .To get the same solutions as of Section 2.1.1, we can consider the case w ≤ , u ≤ w = 1 , u = 66316. Then w = 132632, v = 862108, w = 431054 and hence = × + × + .Let p = 1726201. For w , u ≤ w = 125 , u = 864. Then w = 256, v = 127904, w = 431676 and hence = × + × + . The last solution is for w = 473 , u = 228. Then w = 1444, v = 2736152, w = 432024 and hence = × + × + .To get the same solutions as of Section 2.1.1, we can consider the case w ≤ , u ≤ w = 15 , u = 7790. Then w = 1025, v = 56785, w = 431566 and hence = × + × + .Thus the Erd˝os-Straus conjecture is true for all values of natural numbers n ≥ Note : Let w ∈ N , then (4 w + 3) u − p +34 − w >
0. Thus v = (4 w + 3) w − u = (4 w +3) u (4 w +3) u − p +34 − w − u = u ( p +34 + w )(4 w +3) u − p +34 − w >
0. Hence w ∈ N implies v ∈ N . For w = 4 w + 3, p = 4( w − w ) − x = u p, y = v p, z = w , u v = w w , u + v = w w .Then w = p +34 + w , u v = w ( p +34 + w ) , u + v = w (4 w + 3). Thus u v u + v = p +34 + w w +3 . Hence u v (4 w + 3) = ( p +34 + w )( u + v ). 13 .2 Corollary If w = 1 and u + v = 4 w −
1, then p = 4 { ( u (4 w − u − − w ) } + 1 and 4 p = 1 u p + 1(4 w − u − p + 1 u (4 w − u − . (10) Proof : Let w = 1. Then from equation (8), we get p = 4 u v − u − v , x = u p, y = v p, z = u v . If u + v = 4 w −
1, then p = 4 { ( u (4 w − u − − w ) } + 1 , y = (4 w − u − p . Thus u p + w − u − p + u (4 w − u − = (4 w − u − u + pu (4 w − u − p = (4 w − { ( u (4 w − u − − w ) } +1 u (4 w − u − p = u (4 w − u − u (4 w − u − p = p .If (a) u = 1, then p = 12 w − w + 5 (b) u = 2, then p = 28 w −
23 = 28 w + 5 (c) u = 3, then p = 44 w −
47 = 12 w + 41 (d) u = 4, then p = 60 w −
79 = 60 w + 41 (e) u = 5,then p = 76 w −
119 = 76 w + 33 (f) u = 6, then p = 92 w −
167 = 12 w + 17 (g) u = 7, then p = 108 w −
223 = 108 w + 101 (h) u = 8, then p = 124 w −
287 = 124 w + 85 (i) u = 9, then p = 140 w −
359 = 140 w + 61 (j) u = 10, then p = 156 w −
439 = 156 w + 29 etc.(II) Let p divides 4 u −
1. Then 4 u − v p, v ∈ N . From (7), we get u ( y + z ) = yzv i.e. v u = y + z .(i) If v divides u , then v divides u and 4 u −
1. Thus v = 1, p = 4 u − u + 3 , x =( u + 1) p and one possible values of y, z are y = ( u + 2) , z = ( u + 1)( u + 2) where u = u + 1.Then 4 p = 1( u + 1) p + 1( u + 2) + 1( u + 1)( u + 2) = 1 p ( p +1)4 + 1 p +14 + 1 + 1 p +14 ( p +14 + 1) . (11)(ii) If v = v + v with v and v divide u , then y = u v , z = u v . Also if for some suitablescalar r ∈ N , y + z = v u = r v r u with r v = v + v such that v and v divide r u , then y = r u v , z = r u v . Acknowledgment:
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