The Greek mathematician Diophantus was born circa the year 200 AD in Alexandria. He is best known for his work to solve indeterminate equations in rational numbers. One of the puzzles he set was to find sets of unequal fractions such that the product of any two is one less than a square. We don't know why he picked this problem but it certainly turns out to be an interesting one. One such set of four which he found was,
1/16, 33/16, 17/4, 105/16
In the seventeenth century the study of diophantine equations was taken up by Fermat who gave us the famous problem of proving Fermat's Last Theorem. Wile's brilliant solution has recently stimulated renewed interest in number theory.
Fremat also looked at Diophantine's problem but he was more interested in whole number solutions than fractions. He found,
1, 3, 8, 120
1*3 = 22-1 1*120 = 112-1 1*8 = 32-1 3*120 = 192-1 3*8 = 52-1 8*120 = 312-1
This example raises the question: "Is there a fifth positive integer whose product with any of the four numbers 1, 3, 8 and 120 is one less than a square?"
Baker and Davenport solved this problem in 1969. It is possible to derive an upper limit for the size of any fifth number. A numerical search was required to complete their proof that no such fifth number exists.
However, there is a fifth number which is rational. Euler found the sequence,
1, 3, 8, 120, 777480 / 28792.
There are also many other 4-tuples with the same property so the next problem is more general, but first some terminology:
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Definition 1: A diophantine n-tuple is a set of n positive integers such that the product of any two is one less than a square integer |
.. and because Diophantus was more interested in fractions...
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Definition 2: A rational diophantine n-tuple is a set of n different positive rational numbers such that the product of any two is one less than a square rational |
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Problem 1: Does there exist a diophantine 5-tuple? |
This problem is as yet unsolved and is the one which I would propose as the successor to Fermat's Last Theorem. My reason is that there seem to be lots of different possible attacks on the problem which lead into some interesting areas but a complete solution seems very difficult.
Richard Pinch has developed a computer program which can quickly determine whther or not there is a fifth number corresponding to any given 4-tuple, but there is no known way to eliminate an infinite class of them.
As a first step it would be useful to find all the diophantine 4-tuples. It is easy enough to find an infinite sequence of them, for example,
x-1, x+1, 4x, 4x(2x-1)(2x+1)
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Problem 2:find a way to construct all the diophantine 4-tuples (a, b, c, d). |
It is easy to find the pairs of numbers (a,b) such that the product is one less than a square. Just take any number x and factorise x2-1 into the product of the two numbers a and b. It is well known that x2-1 = (x-1)(x+1) which gives one special solution, and there are many others.
It is also always easy to add a third number c to
the set. Suppose ac+1 = y2. We also need
bc+1 = z2 to be a square, but
b = (x2-1)/a and
c = (y2-1)/a.
Therefore,
bc+1 = (x2-1)(y2-1)/a2
Notice that this is a little less than the square of xy/a,
so it may be useful to ask "how mauch less?" I.e.
what is w if
(x2-1)(y2-1)/a2 = ((xy-w)/a)2-1
Expanding and simplifying gives,
2xwy - x2 - y2 - z2 + a2 + 1 = 0
This equation is symmetric in x, y and w. Think about what this implies! If we have found a solution to the equation then we have a Diophantine 3-tuple, but we can also swap around x, y and w, which means that the number d = (w2-1)/a can be added to give a diophantine 4-tuple.
In 1978 I found that 4-tuples constructed in this way were solutions of a special diophantine equation:
(a + b - c - d)2 = 4(ab+1)(cd+1)
When this equation is expanded it is symmetrical in the four variables so it is also equivalent to,
(a + c - b - d)2 = 4(ac+1)(bd+1) (a + d - c - b)2 = 4(ad+1)(cb+1)
If we look at it as an equation in any one of the four variables (say d) with the other three fixed, it is quadratic with coefficient of one on the square.
d2 + nd + m = 0
If such a quadratic has one integer root d then the other root is, d' = n - d which must also be an integer. If the roots are distinct and positive this can be used as a way to generate a new solution (a, b, c, d'). If the original 4-tuple had the desired property that the product of any two is one less than a square then from the forms of the equation above it is clear that the new one must too. By using this replacement trick repeatedly for different chioces of the four variables we can hope to generate lots of solutions.
If we explicitly complete the square on the 4-tuple equation the discriminate conveniently factorises giving us
(d - a - b - c - 2abc)2 = 4(ab+1)(ac+1)(bc+1)
This gives us a proof of the following theorem
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Theorem 1 For any diophantine 3-tuple (a, b, c), there is fourth positive integer d greater than any of the first three such that (a,b,c,d) is a diophantine 4-tuple It is given by the formula d = a + b + c + 2abc + 2sqrt[(ab+1)(ac+1)(bc+1)] |
This is a solution of the 4-tuple equation and it follows from the given forms above that the required properties hold. The value of d is evidently positive and d > 4abc which is larger than any of them. There is a second root but the negative sign means that it may not be positive. If the original triple is (1,3,8) then the two roots are 0 and 120 but in many other cases both are positive.
Another useful theorem is the next
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Theorem 2 If the 4-tuple of non-negative integers (a,b,c,d) is a solution of the diophantine equation, (a + b - c - d)2 = 4(ab+1)(cd+1) then the product of any two is one less than a square. |
This is not immediately obvious and the proof is by infinite descent. Suppose without loss of generality that 0 <= a <= b <= c <= d. We can generate 4 other solutions in integers by substituting any of the four variables by the other root of the quadratic equation in that variable. E.g. we can replace d, which by,
d' = 4abc + 2a + 2b + 2c - d
This is a distinct solution in integers since,
(d'-d)2 = 16(ab+1)(bc+1)(ac+1)
which is not zero. But d' could be negative, then the 4-tuple (a,b,c,d') would be a solution for which a, b, c are non-negative and d' is negative but from
( a + b - c - d' )2 = 4(ab+1)(cd'+1)
The right is non-negative which implies cd >= -1, so either c = 0 or cd' = -1. If cd' = -1 then d'=-1, c=1 then a = b = 0. If c=0, then similarly a = b = 0 because they can not be larger than c then d' = -2.
So there are exactly two solutions in integers for which a, b and c are non-negative and d is negative, namely (0,0,0,-2) and (0,0,1,-1). In all other case replacing d by d' gives another solution in non-negative integers
Next we can show that d' is less than c
dd' = (c - a - b)2 - 4(ab+1) < c2
( given that c >= b >= a >= 0)
so d >= c => d' < c.
Now we have done enough to show that given a solution (a, b, c, d) we can repeatedly substitude the greatest of the four numbers by its other root giving a smaller solution each time until we get down to one of the two solutions where one of the four numbers is negative, i.e. (0,0,0,-2) or (0,0,1,-1). These two solutions certainly have the property that the product of any two is one less than a square.
To complete the proof we need to show that ( a,b,c,d ) has the property that the product of any two is one less than a square if the same is true for ( a,b,c,d' ). This is easy to get from,
( a + b - c - d )2 = 4(ab+1)(cd+1)
and the similar equations with a, b, c permuted. This completes the proof by induction that any solution in non-negative integers of the 4-tuple equation has the desired property.
We now also know that from any one solution in non-negative integers we can generate a tree of solutions by repeatedly replacing any of the four numbers by the other root of the equation as a quadratic in that variable leaving the other three fixed. There are exactly two trees of solutions, the odd tree which included the solution (0,0,1,3) and the even tree which includes (0,0,0,2).
It is likely that the converse for positive integers is also true i.e.
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Definition 3: A regular diophantine 4-tuple is a solution in positive integers of the equation (a + b - c - d)2 = 4(ab+1)(cd+1) |
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Conjecture 1: Any 4-tuple of positive integers for which the product of any two is one less than a square is a solution of the equation,
( a + b - c - d )2 = 4(ab+1)(cd+1)
In other words, All diophantine 4-tuples are regular.
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I proposed this conjecture in 1978 at about the same time as Arkin, Hoggatt and Strauss made equivalent conjectures. If this conjecture were true then problem 2 would be solved since we can construct all non-negative solutions of the equation. The conjecture also implies that there are no 5-tuples of positive integers (a, b, c, d, e) with a < b < c < d < e for which the product of any two is one less than a square. If there were the two 4-tuples which leave out d and e would be solutions of the equation, so d and e would be the two roots for fixed a, b, c, but as we saw above, the two roots cannot be both bigger than c. So the answer to the main problem 1 would be "no".
The origins of the magic 4-tuple equation seem rather mysterious. There is no explanation for the identities which are used in the above proofs. Perhaps a better understanding may be found by looking at the way the numbers factorise. Take the first two numbers,
ab = x2 - 1 = (x - 1)(x + 1)
It must be possible to factorise a and b such that
a = pq
b = st
x - 1 = ps
x + 1 = qt
=> qt - ps = 2
This factorisation will not be unique if p and t have a common factor or s and q have a common factor but let's not worry about that for the moment.
Similar factorisations for the products bc and ac exist and these can be unified into factorisation of 4 factors for each number such that,
a = a1 a2 a3 a4
b = b1 b2 b3 b4
c = c1 c2 c3 c4
a1 a2 b3 b4 - b1 b2 a3 a4 = 2
a1 c2 a3 c4 - c1 a2 c3 a4 = 2
b1 c2 c3 b4 - c1 b2 b3 c4 = 2
We know that the 4-tuple equation can be applied to these numbers to get the fourth number d. It is interesting to do so and see how it factorises. The answer is that we get,
d1 = b2 a3 c4 + a2 c3 b4
d2 = b1 c3 a4 + a1 b3 c4
d3 = a1 c2 b4 + c1 b2 a4
d4 = c1 a2 b3 + b1 c2 a3
d = d1 d2 d3 d4/4
From this definition of the factors of d we can derive the following:
c3 c4 d3 d4 - c1 c2 d1 d2 = ( a1 c2 a3 c4 - c1 a2 c3 a4 )( b1 c2 c3 b4 - c1 b2 b3 c4 ) = 4 a2 a3 d2 d3 - a1 a4 d1 d4 = ( a1 c2 a3 c4 - c1 a2 c3 a4 )( a1 a2 b3 b4 - b1 b2 a3 a4 ) = 4 b1 b3 d1 d3 - b2 b4 d2 d4 = ( b1 c2 c3 b4 - c1 b2 b3 c4 )( a1 a2 b3 b4 - b1 b2 a3 a4 ) = 4 b1 c3 d3 - c1 b2 d2 = a1 ( b1 c2 c3 b4 - c1 b2 b3 c4 ) = 2 a1 b1 c2 d1 - c4 b2 d4 = a2 ( b1 c2 c3 b4 - c1 b2 b3 c4 ) = 2 a2 b4 c3 d4 - c1 b3 d1 = a3 ( b1 c2 c3 b4 - c1 b2 b3 c4 ) = 2 a3 b4 c2 d2 - c4 b3 d3 = a4 ( b1 c2 c3 b4 - c1 b2 b3 c4 ) = 2 a4
Although the symmetry of these equations of obscured it is not difficult to check that they can be rearranged into the same forms with the roles of a, b, c, d permuted
Now we turn to some interesting peoperties of the 4-variable polynomial whos integer roots are our 4-tuples
P(a,b,c,d) = a2 + b2 + c2 + d2 - 2ab - 2ac - 2ad - 2bc - 2cd - 4abcd - 4
We have already used the fact that,
4(ab+1)(cd+1) = (a + b - c - d)2 - P(a,b,c,d) 4(ac+1)(bd+1) = (a + c - b - d)2 - P(a,b,c,d) 4(ad+1)(cb+1) = (a + d - c - b)2 - P(a,b,c,d) 4(db+1)(dc+1)(bc+1) = (a - b - c - d - 2bcd)2 - P(a,b,c,d) 4(da+1)(dc+1)(ac+1) = (b - a - c - d - 2acd)2 - P(a,b,c,d) 4(da+1)(db+1)(ab+1) = (c - a - b - d - 2abd)2 - P(a,b,c,d) 4(ab+1)(ac+1)(bc+1) = (d - a - b - c - 2abc)2 - P(a,b,c,d)
These relations are curious enough but the mystery deepens further when we find that,
4(da+1)(ba+1)(ca+1) = (a2 - da - ba - ca - 2)2 - a2P(a,b,c,d) 4(ab+1)(db+1)(cb+1) = (b2 - ab - bd - cb - 2)2 - b2P(a,b,c,d) 4(ac+1)(bc+1)(dc+1) = (c2 - ac - bc - cd - 2)2 - c2P(a,b,c,d) 4(ad+1)(bd+1)(cd+1) = (d2 - ad - bd - cd - 2)2 - d2P(a,b,c,d)
Each of these identities takes the form,
M(a,b,c,d) = Y(a,b,c,d)2 - X(a,b,c,d)2P(a,b,c,d)
Where X and Y are simple polynomials and M is a product of the squares which are the key to the puzzle. The set of equations can be further enlarged by using a combination rule,
M1 = Y12 - X12P
and
M2 = Y22 - X22P
=>
M1M2 = (Y1Y2 - X1X2P)2 - (X1Y2 - X2Y1)2P
and
M1M2 = (Y1Y2 + X1X2P)2 - (X1Y2 + X2Y1)2P
One of the results we can get is,
4(ab+1)(ac+1)(ad+1)(bc+1)(bd+1)(cd+1) =
(abcd{a + b + c + d} + 2abc + 2abd + 2bcd + 2acd + a + b + c + d)2 -
(abcd-1)2P(a,b,c,d)
In general we can establish the following theorem,
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Theorem 3 Any product M(a,b,c,d) of 4 times polynomials which are one plus the product of two of the four variables (ab+1 etc), and in which each variable appears an even number of times or each variable appears an odd number of times, is identically equal to a polynomial expression of the form, M(a,b,c,d) = Y(a,b,c,d)2 - X(a,b,c,d)2P(a,b,c,d) |
This raises the question "Is there a polynomial in five variables P(a,b,c,d,e) which has similar properties?" It is not too difficult to find that the answer is yes given that P(a,b,c,d,0) = P(a,b,c,d). The solution is,
P(a,b,c,d,e) = {abcde}2 - 2abcde{a+b+c+d+e} + a2 + b2 + c2 + d2 + e2
- 4abcd - 4abce - 4abde - 4acde - 4bcde - 4
- 2ab - 2ac - 2ad - 2ae - 2bc - 2bd - 2be - 2cd - 2ce - 2de
A few of the identities which this equation satisfies are,
4(ab+1)(ac+1)(bc+1)(de+1) =
(abcde + 2abc + a + b + c - d - e)2
- P(a,b,c,d,e)
4(ae+1)(be+1)(ce+1)(de+1) =
(e2 - ae - be - ce - de - 2 + abcde2)2
- e2P(a,b,c,d,e)
4(ab+1)(ac+1)(ad+1)(bc+1)(bd+1)(cd+1) =
(e{abcd-1}2 - abcd{a + b + c + d}
- 2abc - 2abd - 2bcd - 2acd - a - b - c - d)2
- (abcd-1)2P(a,b,c,d,e)
Combining the last two gives,
4(ab+1)(ac+1)(ad+1)(bc+1)(bd+1)(cd+1)(ae+1)(be+1)(ce+1)(de+1) =
Y2
- ({abcde}2 - abcde{a+b+c+d+e} - abcd - abce - abde - acde - bcde + 1)2P(a,b,c,d,e)
The polynomial Y according to maple is,
b + a + e + c + d + 6 a b c d e + 2 a b c + 2 a b d + 2 b c d + 2 a c d
3
+ 2 e a b + 2 e a c + 2 e a d + 2 e b c + 2 e b d + 2 e c d + a b c d e
2 2 2 2 3 2 2 2 2
- 2 a b c d e + a b c e + a b d e + a c d e + b c d e
3 3 3 3 3 2 2 2 2 2 2 2 2 2 2 2 2
+ a b c d e - 3 a b c d e - 3 a b c d e - 3 a b c d e
2 2 2 2 3 2 2 2 2 2 3 2 2 2 2 2 3 2 2
- 3 a b c d e - 2 a b c d e - 2 a b c d e - 2 a b c d e
2 2 2 3 2 3 3 3 3
- 2 a b c d e + e a b c d + e a b c d + e a b c d + e a b c d
2 2 2 2 2 2 2 2 2
- 3 e a b c d + a b c d + a b c d + a b c d + a b c d + a b c e
2 2 2 2 2 2
+ a b c e + a b c e + a b d e + a b d e + a b d e + b c d e
2 2 2 2 2
+ b c d e + b c d e + a c d e + a c d e + a c d e
As I said before, in the eighteenth century Leonhard Euler found that the sequence 1, 3, 8, 120 has a fifth term which is rational:
777480
1, 3, 8, 120, ------
28792
Now I can state a general theorem,
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Theorem 4 Given any diophantine 4-tuple (a,b,c,d), there is a positive rational number e such that (a,b,c,d,e) is a rational diophantine 5-tuple. It is given by the formula,
e = [{abcd + 1}{a + b + c + d} + 2abc + 2abd + 2bcd + 2acd
+ 2sqrt{(ab+1)(ac+1)(ad+1)(bc+1)(bd+1)(cd+1)}]/(abcd-1)2
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If conjecture 1 above is true then the expression simplifies to,
e = 4sqrt{(ab+1)(ac+1)(ad+1)(bc+1)(bd+1)(cd+1)}/(abcd-1)2
The next step should be to find the polynomial in six variables and continue the sequence.
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Problem 3: Is there a polynomial in six variables P(a,b,c,d,e,f) with properties analogous to the four and five variable polynomials? |
Sadly, this is not so easy to solve. The methods which I used to get from the 4 variable polynomial to the 5 variable one do not work again. If the 6 variable polynomial exists at all it looks like it will have terms including at least the fourth powers of the variables. In that case it will probably not be possible to solve it by the same recursive methods that work when it has only squares. For the moment I can only set a few more unsolved problems.
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Problem 4: Are there any sequences of six or more different non-zero rational numbers such that the product of any two is one less than a square? |
If the word "different" had not been included then the problem would be easy to solve. Just repeat three-quarters as many times as you like!
There are many other related problems which could be set but perhaps the most interesting of them is to explain where the polynomial P(a,b,c,d,e) comes from and why it has the strange properties it does. Perhaps if that was known then some of the other problems might be soluable. If anyone finds an answer please let me know!
I have learnt that there has been some literature on it in maths journals. For a long bibliography and his own substantial work visit the site of Andrej Dujella.
The problems of determining whether or not non-regular diophantine 4-tuples exist is a difficult one. One approach is to reduce the problem to one concerning regular 4-tuples which might be easier to attack. Using theorem 1 we know that we can construct regular 4-tuples from a non-reular 4-tuple by keeping any three of them and looking for the roots of the 4-tuple equation in the other. By combining these we find that if a non-regular 4-tuple (a,b,c,d) exists with d > c > b > a > 0, then we can construct a system of three regular 4-tuples with the same largest element and with each pair sharing one of a, b or c. It can easily be shown that the introduced elements are not zero. This gives us a useful theorem:
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Theorem 5 An irregular diophantine 4-tuple exists if and only if there is a system of three different regular diophantine 4-tuples in the form,
(a,b,c',d)
(a,b',c,d)
(a',b,c,d)
with d > a,b,c,a',b',c' > 0
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First we should ask if any such pair of diophantine 4-tuples exist, or even if there are two diophantine 4-tuples with the same largest element.
Before wasting too much time trying to prove they don't, it is wise to do a search with a computer. It turns out that these pairs are rare but they do exist. I have found three examples after a search of all 4-tuples with d<1019:
1 16640 16899 1124864520
1 399 703920 1124864520
120 145673 154155 10778986831096
8 120 2805566778 10778986831096
528 3253511 3336933 22929452257545400
15 528 723737525421 22929452257545400
From each of these we can extract four 4-tuples which are nearly irregular diophantine 4-tuples. E.g:
1 399 16640 1124864520
This fails to be a diophantine 4-tuple only because bc+1 is not a square. It may be useful to understand the properties of these in general in order to see if there is any reason why the last square cannot be made.
Suppose you were given the two numbers d = 1124864520 and a = 1 and you were asked to find numbers b<d such that ab+1 = x2 and db+1 = y2. You would have to find solutions of the equation
d x2 - a y2 = d-a
Equations of this sort are known as Pellian equations and
a great deal is known about their solutions. In this case there
are six with b<d. i.e.
b = 0, 399, 16640, 16899, 703920, or 1124797443
These correspond to our two coupled diophantine 4-tuples and a
third with a zero in it.
Diophantus was more interested in rational solutions to his problem than integer ones. Most low denominator rational diophantine 4-tuples are regular but irregular ones are important because they may help us understand why we can't find irregular non-rational examples. The smallest denominator irregular rational diophantine 4-tuple is,
1/4 5 33/4 105/4
These days, a computer can rapidly find many rational diophantine 5-tuples to add to the one Euler found. The lowest known common denominator is 16 in these two examples,
5/16 21/16 4 285/16 420
1/16 33/16 105/16 20 1140
In January 1999 I found the first rational Diophantine 6-tuple:
11/192 35/192 155/27 512/27 1235/48 180873/16
This page was last updated 17 January 1999.
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