On the Diophantine Equation 2^a3^b + 2^c3^d = 2^e3^f + 2^g3^h
aa r X i v : . [ m a t h . G M ] O c t On the Diophantine Equation 2 a b + 2 c d = 2 e f + 2 g h Roger TianNovember 21, 2018
Abstract
This paper is a continuation of [1], in which I studied Harvey Friedman’s prob-lem of whether the function f ( x, y ) = x + y satisfies any identities; however, noknowledge of [1] is necessary to understand this paper. We will break the exponen-tial Diophantine equation 2 a b + 2 c d = 2 e f + 2 g h into subcases that are easierto analyze. Then we will solve an equation obtained by imposing a restriction onone of these subcases, after which we will solve a generalization of this equation. Acknowledgements
I would like to thank my thesis advisor, George Bergman, for giving advice onhow to better organize this paper and make it more readable, and for pointing outareas of my paper that needed clarification.
We will follow the convention that 0 / ∈ N .A resticted version of Friedman’s problem (mentioned in the Abstract) I studied inmy paper [1] is related to the solution set of the equation2 a b + 2 c d = 2 e f + 2 g h . (1)The results we will get on this equation, which are still very partial, will not be appliedin this paper to Friedman’s problem.Suppose a , b , c , d , e , f , g , h are nonnegative integers such that2 a b + 2 c d = 2 e f + 2 g h . (2)Without loss of generality, we may assume that min { a , c , e , g } = 0 =min { b , d , f , h } , or equivalently 0 ∈ { a , c , e , g } and 0 ∈ { b , d , f , h } ; we canalways divide (2) by 2 min { a ,c ,e ,g } min { b ,d ,f ,h } . Suppose that there is exactly onezero in { a , c , e , g } . Then (2) reduces to an equation in which exactly three of its fourterms contain a factor of 2, so one side of this resulting equation is divisible by 2 whilethe other side is not, which is a contradiction. Thus, there must be at least two zerosin { a , c , e , g } and, by the same reasoning, with the factor 2 replaced by the factor3, there must be at least two zeros in { b , d , f , h } . Then, depending on which termsof (2) the zeros occur in, we can reduce Equation (1) to 36 cases. However, merging1he cases that are identical up to permutations of the summands, we get the followingseven equations: 1 + 1 = 2 e f + 2 g h (3)1 + 3 d = 2 e + 2 g h (4)3 b + 3 d = 2 e + 2 g (5)1 + 2 c = 3 f + 2 g h (6)1 + 2 c d = 1 + 2 g h (7)3 b + 2 c = 1 + 2 g h (8)3 b + 2 c = 3 f + 2 g (9)Note, for instance, that in the case a = b = c = f = 0, Equation (4) must have atleast one solution. The solution to (3) is ( e, f, g, h ) = (0 , , , c, d, g, h ) = ( s, t, s, t ) for all nonnegative integers s and t . We now solve (8) subjectto the restiction b = h , i.e. the equation 2 c − b (2 g − Lemma 1.
Let p, m, n ∈ N ∪ { } where p > and m > . If p m − | p n − , then m | n .Proof. We have n = qm + r where q, r ∈ N ∪ { } and 0 ≤ r < m . We will prove thislemma by induction on q . For q = 0, we have n < m , so p n − < p m −
1, from which itfollows that p m − | p n − ⇒ p n − ⇒ n = 0 = ⇒ m | n . Suppose the lemmais true for some q , we will prove it for q + 1. Suppose p m − | p n − p ( q +1) m + r − p m − p ( q +1) m + r − − ( p m −
1) = p ( q +1) m + r − p m = p m ( p qm + r − p m − p m are relatively prime, we have p m − | p qm + r −
1. It follows from ourinductive hypothesis that m | qm + r , so r = 0 and n = ( q + 1) m . This completes theinduction. Notation . Let n, m, k ∈ N . By n k k m we will always mean that n k | m and n k +1 ∤ m . Lemma 3. If k, n ∈ N where k is odd, then n +2 k n k − .Proof. We will first prove this claim for n = 1. We have 3 k − k − k − − P kj =0 (cid:0) kj (cid:1) j = P kj =1 (cid:0) kj (cid:1) j = 8 P kj =1 (cid:0) kj (cid:1) j − , and we see that P kj =1 (cid:0) kj (cid:1) j − is oddbecause the j = 1 term of this sum is odd and all other terms of this sum are even.Now suppose the claim is true for some n ≥
1. Then there exists an l ∈ N such that l is odd and 3 n +1 k − n k ) − n k − − n +2 l + 1) − n +2) l + 2(2 n +2 l ) + 1 − n +2) l + 2(2 n +2 l ) = 2(2 n +2 l )(2 n +1 l + 1), and we seethat 2 n +1 l + 1 is odd and 2 n +3 k n +1 k −
1. This completes the induction.
Lemma 4. If m is odd, then k m + 1 .Proof. Notice that for m = 1 we have 3 m + 1 = 3 + 1 = 4. Now suppose that theclaim is true for some m ≥
1, where m is odd. We have 3 m +2 + 1 = 9 · m + 1 =9(3 m + 1 −
1) + 1 = 9(4 k −
1) + 1 = 36 k − k − k −
2) where k is odd,and we see that 9 k − emma 5. If m , l ∈ N and m ∈ N ∪ { } where , ∤ l , then m +1 k m m l − .Proof. We will first prove this claim for m = 0. Notice that 2 m l − m − l − m − l − P m − li =1 (cid:0) m − li (cid:1) i +1 − P m − li =1 (cid:0) m − li (cid:1) i = P m − li =2 (cid:0) m − li (cid:1) i +2 m − l · P m − li =2 (cid:0) m − li (cid:1) i + 2 m − l ), and we see that 3 ∤ P m − li =2 (cid:0) m − li (cid:1) i +2 m − l . Now suppose for some m ≥ m m l − m +1 l where 3 ∤ l . Then2 m m l = 3 m +1 l + 1 = ⇒ m m l = (3 m +1 l + 1) = 3 m +1) l + 3 · m +1) l +3 · m +1 l + 1, so 2 m m l − · m +1 l (3 m +1) − l + 3 m +1 l + 1). Since3 ∤ l (3 m +1) − l + 3 m +1 l + 1), we have 3 m +2 k m m l −
1. This completes theinduction.
Lemma 6. If m , l ∈ N ∪ { } where , ∤ l , then m +1 k m l + 1 .Proof. We will first prove this claim for m = 0. Notice that 2 l + 1 = (3 − l + 1 = P lk =1 (cid:0) lk (cid:1) k ( − l − k + ( − l + 1 = P lk =1 (cid:0) lk (cid:1) k ( − l − k − P lk =1 (cid:0) lk (cid:1) k ( − l − k = 3 P lk =1 (cid:0) lk (cid:1) k − ( − l − k = 3( P lk =2 (cid:0) lk (cid:1) k − ( − l − k + l ), and we seethat 3 ∤ P lk =2 (cid:0) lk (cid:1) k − ( − l − k + l . Suppose for some m ≥ m l + 1 = 3 m +1 l where 3 ∤ l . Then we have 2 m l = 3 m +1 l − ⇒ m l = (3 m +1 l − =3 m +1) l − · m +1) l + 3 · m +1 l −
1, so 2 m l + 1 = 3 · m +1 l (3 m +1) − l − m +1 l + 1). Since 3 ∤ l (3 m +1) − l − m +1 l + 1), we have 3 m +2 k m l + 1. Thiscompletes the induction. Proposition 7.
The only solutions ( k, m, n ) in the positive integers of the exponentialDiophantine equation k (2 m −
1) = 2 n − are (1 , , and (2 , , .Proof. We know from Lemma 1 that n = lm for some l ∈ N . Since k >
0, we must have l ≥
2. Notice that 2 n − lm − m − ( l − m +2 ( l − m + . . . +2 m +1) = 3 k (2 m − k = 2 ( l − m + 2 ( l − m + . . . + 2 m + 1 > m − . (10)It follows that 3 k > k (2 m − k > n − . (11)Now, by Lemma 5 we have 3 k | n − ⇒ n = 2 m m l where m , l ∈ N and m ∈ N ∪ { } such that 2 , ∤ l and m ≥ k −
1. Note that m = k − k k n − k < m k − l − k ≥ m and l . It is easy to check that, for all choices of m and l , wehave 3 k < m k − l − k = 3. Suppose we know, for some value of k ≥ k < m k − l −
1, or equivalently 3 k + 1 < m k − l , for all choices of m and l . Then we have 2 m k l = (2 m k − l ) > (3 k + 1) > k +1) + 1 for allchoices of m and l . This completes the induction. If k ≥ k | n −
1, then2 n − m m l − ≥ m k − l − > k , contradicting (11). Thus, we see thatthere are no solutions for k ≥ k = 1 ,
2. Suppose k = 1. Then3 k = 3 > n − n = 1, 2, or 3, so we must have n = 2 because n is even,3ence m = 1. Suppose k = 2. Then by (10) we have 3 = 9 > m −
1, so m = 1, 2, or 3.It is easy to check that the cases m = 1 and m = 2 do not yield solutions. For m = 3,we have n = 6. Remark . Central to our proof is the fact that, for k sufficiently large, 2 n − k · q implies q must be much larger than 3 k .We will now solve a more general exponential Diophantine equation using the sameideas in the proof of the preceding proposition. First, we generalize Lemma 5. Lemma 9.
Let m ≥ be an odd positive integer. Then the following statements hold.1. If n ∈ N is odd, then m ∤ ( m − n − .2. If m , l ∈ N , m ∈ N ∪ { } , and , m ∤ l , then m m +1 k ( m − m m m l − .Proof. We consider the two cases separately.1. We have ( m − n − ≡ ( − n − ≡ − − ≡ − m ), from which theconclusion follows.2. Our proof is by induction on m .For m = 0, we have ( m − m l − P m li =0 (cid:0) m li (cid:1) m i ( − m l − i − − P m li =1 (cid:0) m li (cid:1) m i ( − m l − i = P m li =1 (cid:0) m li (cid:1) m i ( − m l − i = m m l X i =2 (cid:18) m li (cid:19) m i − ( − m l − i ! − m l ! . (12)Notice that m | P m li =2 (cid:0) m li (cid:1) m i − ( − m l − i , but m ∤ m l because m ∤ l andgcd( m, m ) = 1. Thus, m ∤ P m li =2 (cid:0) m li (cid:1) m i − ( − m l − i − m l and so m k ( m − m l − m ≥ m − m m m l − m m +1 l where m ∤ l . Then ( m − m m m l = m m +1 l + 1 = ⇒ ( m − m m m l =( m m +1 l + 1) m = P mi =0 (cid:0) mi (cid:1) ( m m +1 l ) i = 1 + P mi =1 (cid:0) mi (cid:1) ( m m +1 l ) i , so ( m − m m m l − P mi =1 (cid:0) mi (cid:1) ( m m +1 l ) i = m · m m +1 l + P mi =2 (cid:0) mi (cid:1) ( m m +1 l ) i = m m +2 l + P mi =2 (cid:0) mi (cid:1) m ( m +1) i l i = m m +2 ( l + P mi =2 (cid:0) mi (cid:1) m ( m +1)( i − − l i ). Byassumption m ∤ l . However, m | P mi =2 (cid:0) mi (cid:1) m ( m +1)( i − − l i because m clearlydivides (cid:0) mi (cid:1) m ( m +1)( i − − l i for all i ≥ (cid:0) m (cid:1) m m l = m ( m − m m l = m − m m +1 l is divisible by m since m − m ∤ l + P mi =2 (cid:0) mi (cid:1) m ( m +1)( i − − l i . This completes the induction. Remark . Part 1 of the lemma holds for all integers m ≥
3. If m is even, then we canwrite m = 2 k p where k, p ∈ N and 2 ∤ p , and we can observe (taking m = k and l = p )from (12) that m | ( m − k p −
1. Thus, Part 2 of the lemma is false for even m .4 roposition 11. The only solutions ( k, p, q, n ) in the positive integers of the exponentialDiophantine equation (2 n + 1) k ((2 n ) p −
1) = (2 n ) q − are (2 , , , and (1 , , , n ) forall positive n .Proof. We know from Lemma 1 that q = lp for some l ∈ N . Since k >
0, we must have l ≥
2. Notice that (2 n ) q − n ) lp − n ) p − n ) ( l − p +(2 n ) ( l − p + . . . +(2 n ) p +1) = (2 n +1) k ((2 n ) p − n +1) k = (2 n ) ( l − p +(2 n ) ( l − p + . . . +(2 n ) p +1 > (2 n ) p −
1. It follows from (2 n + 1) k > (2 n ) p − n + 1) k > (2 n + 1) k ((2 n ) p − n + 1) k > (2 n ) q − . (14)Since (2 n +1) k | (2 n ) q −
1, by Lemma 9 we have q = 2 m (2 n +1) m l where m , l ∈ N such that 2 , n + 1 ∤ l and m ∈ N ∪ { } such that m ≥ k − k ≥ n + 1) k < (2 n ) m (2 n +1) k − l − n , m , and l . For k = 3, observe that(2 n + 1) k < (2 n ) m (2 n +1) k − l − n , m , and l ; take x := 2 n + 1, m = 1 = l and note that x < x x − < (( x − ) x − x − x − x ≥ k ≥
3, that (2 n + 1) k < (2 n ) m (2 n +1) k − l − ⇐⇒ (2 n + 1) k + 1 < (2 n ) m (2 n +1) k − l for all choices of n , m , and l . Then(2 n ) m (2 n +1) k l > ((2 n + 1) k + 1) n +1 ≥ (2 n + 1) k (2 n +1) + 1 ≥ (2 n + 1) k +1) + 1 for allchoices of n , m , and l . This completes the induction. If k ≥ n +1) k | (2 n ) q − n ) q − n ) m (2 n +1) m l − ≥ (2 n ) m (2 n +1) k − l − > (2 n + 1) k ,contradicting (14). Thus, there are no solutions for k ≥ k = 1 ,
2. Consider k = 1. Thenby (13) we have 2 n + 1 > (2 n ) p − , (15)which for n = 1 becomes 4 > p , which is false for every p ≥
2. Notice that as n increases, (2 n ) p − p ≥ n + 1. It follows that (15) isfalse for every p ≥
2, so we must have p = 1. For p = 1, we have (2 n + 1) k ((2 n ) p −
1) =(2 n + 1)(2 n −
1) = (2 n ) − n ) q − ⇒ q = 2, and n can be any positive integer.Consider k = 2. Then by (13) we have(2 n + 1) > (2 n ) p − , (16)which for n = 1 becomes 9 > p −
1, which is false for every p ≥
4. Notice that as n increases, (2 n ) p − p ≥ n + 1) . It follows that (16) isfalse for every p ≥
4, so we must have p = 1, 2, or 3. If p = 3, then by (16) we must have n = 1, so 3 (2 −
1) = 2 q − ⇒ q = 6. Take x := 2 n for simplicity of notation as wecheck the remaining two cases. Suppose p = 2. Then we have ( x + 1) ( x −
1) = x q − ⇒ x + 2 x − x − x q − ⇒ x + 2 x − x = x q , so q ≥
5. However, it is easyto see that x + 2 x − x < x q for q ≥
5, so there are no solutions for p = 2. Suppose p = 1. Then ( x + 1) ( x −
1) = x q − ⇒ x + x − x − x q − ⇒ x + x − x = x q ,5o q ≥
4. However, it is easy to see that x + x − x < x q for q ≥
4, so there are nosolutions for p = 1. Notation . For all n, m ∈ N where n ≥ m ≥ k ∈ N ∪ { } suchthat m k k n , and we will denote this k by v m ( n ).Since Lemma 9 allowed us to solve the Diophantine equation of Proposition 11, anatural question is whether this lemma can be extended to all positive integers m ≥ m + 1) k ( m p −
1) = m q − m > Conjecture 13.
There exists a positive integer N such that for all integers n ≥ N andall even integers m > we have ( m v m (( m − n − ) ≤ ( m − n − . References [1] Roger Tian,
Identities of the Function f ( x, y ) = x + y3