THE DIOPHANTINE EQUATION x 4 +1=Dy 2

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1 MATHEMATICS OF COMPUTATION Volume 66, Number 9, July 997, Pages S )0085-X THE DIOPHANTINE EQUATION x 4 +=Dy J. H. E. COHN Abstract. An effective method is derived for solving the equation of the title in ositive integers x and y for given D comletely, and is carried out for all D< If D is of the form m 4 +, then there is the solution x = m, y = ; in the above range, excet for D = 7058 with solution x = 6, y = 57, there are no other solutions. Over fifty years ago, Ljunggren [], showed that the equation of the title, where without loss of generality D is square free, has at most one solution in ositive integers. The method was urely algebraic, but rather comlicated. It is the object of this note to rovide an effective comutational method of determining whether a solution exists and of finding it when it does. Clearly, there can be no solution unless the equation v Du = has solutions, and in articular q mod 4) for every odd rime factor q of D. Letα=a+b D be its fundamental solution, and define β = a b D,and u n = αn β n D ; v n= αn +β n ). Then for the equation of title we have for some odd m, x = v m and y = u m.since b u m we see that for any odd rime factor q of Db, x ±) ±x mod q) and so q mod 8). Since D was assumed square free and clearly b must be odd, we see that b mod8).ifdis odd, then D mod8),andso4 awhence D mod 6). So we have two cases: Case. D mod 6), every factor of D is congruent to modulo 8; Case. D mod 6), every odd factor of D is congruent to modulo 8. In Case we find 4 a and in Case a ±mod8). Theorem. Let a = AB where A is square free. Then m = A rovides the only ossible solution. Proof. Since α + β =a,αβ =, the sequence {v n } satisfies the recurrence ) v n+ =av n+ + v n with initial values v 0 =andv =a. For n odd, let w n = v n /a, which is also an integer. Then by ) w n+4 w n+ = v n+3, w n+ w n =v n+.thus w n+4 w n+ + w n =v n+3 v n+ )=4av n+ 0 mod 4a), Received by the editor March 4, Mathematics Subject Classification. Primary D c 997 American Mathematical Society

2 348 J. H. E. COHN and so since w =,w 3 =3 mod4a), it follows that for all odd n 3) w n n mod 4a). In articular, solutions are ossible only if a = α a where α 0anda is odd. In this case, we have for all odd n with a, n) = 4) w n a )=n a )=a n)ξ=a n)ξ, where ξ = ifn a 3mod4)andξ= otherwise. Next we rove by induction on nn that for all odd corime integers n, N the Legendre-Jacobi symbol w n w N ) = n N). This holds if nn = ; suose it is true for all values less than the one we consider. As n and N are suosed corime, n = N is imossible unless n = N = ; without loss of generality we may assume n>n, since by 3) quadratic recirocity gives w n w N )=w N w n )θ, where θ = ifn N 3mod4)andθ= otherwise. Then it is easily found that w n w n N mod w N ), and again n N and N are corime. If here n N is ositive, then w n w N )=w n N w N )=n N N)=n N) and the induction is comlete; on the other hand if n N is negative, then we use w n N = w N n and then w n w N )=w n N w N )= w N n w N )= w N )N n) N) =n N) in view of 3), since if N<n<N,then0<N n<n. Suose that x = v m = aw m ;letndenote any odd integer corime to am. Then =aw m w n )=a w n )m n) =a w n )m n) =w n a )m n)ξ =am n), by 3) and 4) and this imlies that am must be a erfect square, since otherwise, we may choose n to be congruent to modulo 4 and also be a quadratic non-residue modulo am. ThusAm must be a erfect square, and to comlete the roof we need to show that m cannot have any squared factor. If m,thenmhas an odd rime factor, saym=, where. Then v m + u m D = α =v +u D) and so x = v ) v r D r u r = v ) v r v +)r =v V, say, where V mod v ). Thus we must have either v = x,v =x or v = x, V = x. The former is imossible by Ljunggren s result. Suose then that v = x. We show that. Forif=K, then we obtain as above x = v K = v K V, say, where now V mod vk ). Since v KV it follows that V and hence that v K = x 3, again imossible by Ljunggren s result. This concludes the roof. For a given D it is therefore in rincile trivial to solve the equation; all that is required is to determine a, calculatev A and test whether it is a square. Unfortunately there are a number of ractical difficulties occasioned by the huge values that occur; for examle, when D = 9744, a has 89 decimal digits. Since the determination of A is nearly of the same order of difficulty as a comlete factorisation of a, we consider how the calculations can be reduced. The first ste in the comutation for any D or mod 6) is to eliminate any that are squares or have an odd rime factor mod8),andthentocalculate

3 THE DIOPHANTINE EQUATION x 4 +=Dy 349 the continued fraction of D u to the end of the first eriod. There is a very efficient algorithm for doing this involving only integer arithmetic. If the eriod length is even, as occurs for examle for D = 34, then the equation v Du = has no solutions, and so the equation of the title has none. Assuming the eriod length n to be odd, if D =[a 0,a,...,a n,a 0 ], we next calculate the convergent a/b corresonding to [a 0,a,...,a n ] which rovides the fundamental solution α of v Du =, as is well nown. Provided multirecision is available, this too can be carried out very efficiently. If here b has a factor congruent to 5 modulo 8 again there will be no solution; this occurs for examle with D = 93 for which b = 6985, but it is not always feasible to use this unless b has a small such factor; it is not worth attemting a comlete factorisation of b in other cases, as will become aarent. As we have seen, a solution is ossible only if x = v A, or equivalently if w A = AX. Often, although it is imracticable to factorise a comletely, we are able to find some of the factors of A. The following result is therefore of assistance. Theorem. A solution can exist only if for every factor of A, w = X,and in Case, every factor of A must be congruent to ± mod8). For if A = l, thenasabove x =v l ) ) l r v l r D r u r = v l ) ) l r v l r v +)r =v V and again V l mod v ). But now l a v and so v = lx, V = lx, and so since A = l is square free, w = lx /AB = X. InCasewehavethataand so also x and v are odd, and so modulo 8, l V + whence l ±mod8). l ) +4 l 4) Theorem 3. There is no solution if 3 A. =+ll ) + 6ll )l )l 3), For as above, this would yield with A =3,v =3x,3x =4v +3,andthen x =x 4 +. An easy descent argument shows that this equation is imossible, for it would require x ± =x 4 3,x =6x 4 4 whence ± =x4 3 3x4 4. Here the lower sign is rejected modulo 3, and the uer sign requires x 3 odd and x 4 even. Thus x 3 +)x 3 ) = 48 x 4) 4 and since x 3 +,48) =, we obtain x 3 + = x4 5, x 3 =4x4 6 and so x4 5 =x Theorem 4. There is no solution if 5 A. For, it would give v =5x,5x =6v4 +0v +5withx 0,whereA=5. Thus 4x =540x 4 +), whence 40x 4 +)=F 3r,where{F n }denotes the Fibonacci sequence, and r is odd. To show that this cannot occur, let {L n } denote the Lucas sequence. Using standard identities for these sequences, we find it would require { 80x 4 = F F 6m L 6m+3 if r =4m+ mod 4), 3r F 3 = F 6m L 6m 3 if r =4m 3 mod 4).

4 350 J. H. E. COHN Now none of the Lucas numbers is divisible by 5, and since F 6m,L 6m±3 )=4we seethatwerequirel 6m±3 to be a square or twice a square. By [], this occurs only for 6m ± 3 = 3. The lower sign gives no value, and the uer gives only x =0. Theorem 5. If 3 a, then ±or ±5 mod 4) for every factor of A. For if A =, thenv amod 3) and we now have x ) a r a +) r ) ) r mod 3) = + i) + i) = π cos 4, whence the result. Theorem 6. If 5 a, then ±mod5)for every factor of A. For now we cannot have a 4 mod 5) since 5 Db. Thus a mod5) and then A ±mod5). IfA= we must have v = x,andthen ± mod 5) would give v ± mod 5), since it is easily found that 5 v,andthen ) x = v r v +)r ± mod 5), which is imossible. We therefore tested whether a was a square; if so the roblem is solved affirmatively. If not, then we attemt to eliminate D as simly as ossible, by attemting factorisations of a, to find A, andofb. This may be difficult, but even when comlete factorisations are imracticable, it may be ossible to find enough information, for examle if b has any factor 5mod8),orifAis divisible by or 3 or 5, or if B is not divisible by 3 but A has a rime factor ±7or± mod 4) or if B is not divisible by 5 but A has a rime factor ± mod 5). If this fails to dismiss D, thenwehavetotestforv A =x. Even if A is nown, the numbers are often far too large for a direct demonstration, and we use congruences. Let q denote any rime; then modulo q the sequence {v n } is eriodic, and with eriod z = zq), say, not exceeding q.thenifr Amod z),v A could be a square only if the congruence v r x mod q) were soluble. It is usually quite easy to find a suitable q for which this fails to hold. If A is unnown but some of its factors are, then we use Theorem instead and test for w = X in the same way. To obtain a flavour of the wor involved, when D = 7393, a is a number of 58 decimal digits, not divisible by 5 but having the unreeated factor 3; thus this case can be eliminated by Theorem 6. When D = 47858, a = 379, b = 7 and since 379 is rime, we must consider ν 379. We find that v 379 is not a square, since if q =, z =4andthenv mod ). Not all values of D succumbed quite so easily. All the solutions we have found arise from A = ; we conjecture that this is always the case, although we are unable to rove it. [The equation x 4 =Dy is more amenable to treatment along these lines.] A roof would eliminate much of the comutation, for then all that would be required would be to test whether a is a erfect square. As we have seen, if A> is a solution, then w = X for every factor of A, and we now that we cannot have = 3 or 5. We accordingly mae the stronger

5 THE DIOPHANTINE EQUATION x 4 +=Dy 35 Conjecture. The equation w = X is satisfied for no odd >and a>0. For our uroses it would suffice to rove this for all odd rime dividing A. References. J. H. E. Cohn, Lucas and Fibonacci numbers and some Diohantine equations, Proc. Glasgow Math. Assoc ), 4 8. MR 3:0. Wilhelm Ljunggren, Einige Sätze über unbestimmte Gleichungen von der Form Ax 4 + Bx + C = Dy, Vid-Aad. Sr. Norse Oslo 94 No. 9. Deartment of Mathematics, Royal Holloway University of London, Egham, Surrey TW0 0EX, United Kingdom address: j.cohn@rhbnc.ac.u