What the sum of two quintic powers of natural numbers can’t be. 

INTRODUCTION In 2020 the author of this blog posted an elementary proof of a result Pierre de Fermat claimed to have found more than 350 years ago.  It had taken the current author twenty-nine years of “bare hands” hard grind, trying to find a simple, easily followed, direct, straightforward way to show that a^p + b^p (where a and b were natural numbers and p was a prime number) could never be the pth power of a natural number. The project had been conceived as a landmark gesture in favour of cultivating the use of reasoning in mathematics… and hence, by implication, of giving reasoning itself a lift. 

However a gesture of this kind achieves nothing if it is not taken seriously.

The response of readers was, to put it mildly, disappointing. As far as the author knows, no one had the curiosity, dilligence or fortitude to follow the reasoning carefully through to the end. On the contrary, some formidable mathematical talents seemed to be bent on finding imagined nit-picking difficulties on pages 1 and 2… The Editor of the Mathematical Gazette and two of his referees were unable to find a flaw in the argument. But the journal turned it down anyway, on the presumption that <<there must be a mistake somewhere!>>.

This blog takes the form of a revisit to the proof, involving the case when p=5. This should make the logic behind the reasoning easier to follow. It is to be hoped that some of those who were flummoxed by the generalised reasoning of the 2020 version, will find the current version more transparent.

The aim of this elementary proof is to show conclusively that the sum of two quintics (natural numbers) can’t be another quintic, i.e. that the following is impossible: 

                         a⁵ + b ⁵ = c⁵ , where a, b, and c are natural numbers. 

We know by ordinary observation, backed by systematic calculations using modern computers, that no case of this equation being satisfied has ever been found.  Both a and b here could take an infinity of different values: but puzzlingly, whatever a and b we choose, it appears that a⁵ + b⁵ never equals another quintic natural number. This begins to look as if it might be a logical fact. Actually we now know, thanks to Andrew Wiles’ advanced, hyper-abstract proof, that it is a logical fact. So it is reasonable to suppose that Fermat did have an elementary proof. 

If so, how did it proceed? This is a search for that amazing, special, elementary, but tragically lost Fermat reasoning. To leave this low-level re-discovery issue unsolved for more than 350 years —which is what has happened— casts about as defeatist a reflection on the reasoning power of our mathematics as it is possible to imagine.   

The search starts with the simple observation that, if one takes two arbitrary natural numbers a and b (both > 1),  the sum of a⁵ + b⁵ is always less than (a+b)⁵. Also (a+1) ⁵ + (b+1)⁵ is less than (a+b+1)⁵.  So is (a+2)⁵ + (b+2)⁵ less than (a+b+2)⁵.   For example, 2⁵ + 3⁵ is less than 4⁵, 3⁵ + 4⁵ is less than 6⁵, and 5⁵ + 6⁵ is less than 100,000. But if one continues to add 1s to both a and b, eventually (say, after adding h 1s), the worm turns, and the combined sum of the two natural quintic powers becomes larger than the single quintic power (a+b+h)⁵. 

We can put this formally as follows:   

For a sufficiently large natural number m:                                            .    . .    .                              .   . .  .                                                   (a+m)⁵ + (b+m)⁵ > (a+b+m)⁵.

In other words the size relationship between the sum of the two LHS terms and the RHS term suddenly flips at a special value of m.  This, incidentally, is not surprising, because when m is large compared with a and b, the LHS is roughly 2m⁵ and the RHS is roughly m⁵.  It may be noted that, given a and b, there is only one special value of m. After the LHS becomes the largest side, it stays that way.

The question now becomes interestingly specific: <<Could there be a case  when the two sides are exactly equal?>>. This is the conjecture we pursue in this proof. The conjecture we are hoping to fault —by finding a contradiction— is that, given a and b, a natural number h exists such that:

(a+h)⁵ + (b+h)⁵ = (a+b+h)⁵   .……………………………………………………………..(1) 

If such a special value of h exists, it will, of course, be determined by a and b.  The aim of this exposition is to show by elementary reasoning that no such natural number h can exist. {If we take any natural numbers x, y and z, it is possible to express them in terms of h=x+y-z, a = z-y, b= z-x.]  

If we were willing to let h take the form of a real number, such an h would be a simple, computable function of a and b. The question we are considering is, whether this h can be a natural number. 

Now when we expand the brackets in equation (1) using the Binomial Theorem we get:

h⁵ = 5ab{a³ + 2a²b +2ab² + b³ +h(4a² +6ab + 4b²)+6h²(a+b)+4h³}………………………………..(2)

and this, after factorising, becomes:

 

h⁵ = 5ab(a+b+2h){a² + ab + b² +2h(a+b) +2h²}…………………………………………………….(3)

If there is a positive integer solution to this equation, there will be three natural numbers (a+h), (b+h) and (a+b+h) which satisfy equation (1).

But (2a+2h), (2b+2h), (2a+2b+2h) will also satisfy equation (1) because we have simply imported a factor 32 into both the LHS and the RHS of (1). Similar remarks apply if we import a factor 243, 1024, 3125…etc. 

So it will help to focus the problem if we limit a + h and b+ h to being co-prime.

We know from (3) above that h⁵, and therefore h, already has all the prime factors of both a and b.

Now if a and b had a common factor (say k), both a+h and b+h would also have the prime factors of k.  So if  (a+h) and (b+h) are to be co-prime, a and b must be co-prime too. 

The expression (a + b + 2h) in (3) will be denoted z. The question now arises whether z is co-prime to a and b. It is only necessary to establish this for a and z because a and b are effectively inter-changeable. Now if a and z have a common factor, subtracting,  b+2h will have it too. But b + 2h cannot have a factor common to a, because h has this and b doesn’t. 

So a and z have no common factor and the same form of argument establishes that b and z have no common factor.

[AN ASIDE  Since z = a+b+2h, equation (3) can be replicated using just the unknowns z, b, h or alternatively z, a, h.

We get

h⁵ = 5bz(z–b-2h){z² – zb – b² -2h(z+b) +2h²}…………………….(3’)

or

h⁵ = 5az(z–a-2h){z² – za -a² -2h(z+a) +2h²}…………………….(3”)

 So there is a kind of symmetry here which is useful later in the exposition. END of ASIDE.]

Next, the question arises whether a, b and z are all co-prime to the curly bracket 

{a² + ab + b² +2h(a+b) +2h²}, which will be denoted v………………………………….(4)

So (3) becomes

        h⁵ = 5abzv……………………………………………………………………………(5)

We already know that a, b and z do have factors in common with h, so we can take h for granted and remove terms with a factor h from the reasoning. If we disregard h, the expression for v reduces to a²+ab+b².  This clearly has no factor in common with a or b, and z reduces to a+b. Does a² + ab + b² have a factor common to a+b?  No, because if it did, (a+b)² – ab would have such a factor: which would imply that ab also had a common factor with either a or b… something already ruled out.  

So the result of this preliminary inquiry into the co-prime status of a, b, z, v is that they are all co-prime to each other.

This tells us that each of a, b, z and v must be a quintic power except that one and only one of them must also have the factor 5⁴. (Otherwise there would be an extra factor 5 on the RHS.)  The one which has the factor 625 may also have the factor 3125 to some power.

So it follows that there are natural numbers, A, B, Z, V such that 

a = k₁A⁵

b = k₂B⁵

z = k₃Z⁵

v = k₄V⁵   where just one of the ks is 5⁴     (i.e. 625) and the others are all 1.

Also taking the 5^th root of each side of equation (5) above:

       h = 5ABZV……………………………………………………….(6)

This is a landmark result.

It may be noted here that one and only one of a, b, z, v must have a factor 625 or this times a power of 5.

Now v is either V⁵ (when one of a,b,z has the factor 5) or 625V⁵ (when it, v, is the unknown with the factor 5).

Could it be either of these things? Probably not: there is no reason to think that the curly bracket expression on the RHS of equation (3) has the form of a fifth power, or a fifth power times 625.  This stands out as the most likely elementary reason why no solution has ever been found for equation (1). The problem, though, is to prove this putative impossibility.

There are two main cases: CASE 1 when v lacks the factor 5 and CASE 2 when v has the factor 5. 

CASE 1 breaks into two sub-CASEs (1.1) when z has the factor 5, and (1.2) when a or b has the factor 5. (a and b are interchangeable, so we shall simply look at the case (1.2) when a has the factor 5.)

We look at the two sub-cases in turn. This is case 1 so v =  V⁵ and recapping: 

             V⁵ = a² + ab + b² +2h(a+b) +2h²……………………………………………………….…..(4)

It is striking that the RHS of (4) contains two quintics, a² and b², or if you prefer A¹⁰ and B¹⁰, while everything else has the factor AB.

So let V = A² + B² + y, ………………………………………………………………………………..(6)  where y is the integer needed to make the RHS total up to V

Now because all the terms of V⁵ (in (4)) apart from a² and b² have the factor AB, it looks as if y will necessarily have the form gAB where g is an integer.

The next step is to test this hypothesis. The test we apply here is to ascertain whether y has the factor A. (Because A and B are inter-changeable,  this will suffice.)

Now once A and B have been chosen, all the other quantities are uniquely determined by the equations we have explored in this exposition.  So y, for example, is equal to V – A² – B² where V is the fifth root of the RHS of (4) and the h in this expression is 5ABZV.

It would be difficult to state an explicit expression for y in terms simply of A and B, but it is evident that putting the A item in this expression as zero will result in the new value of y , to be denoted y’, becoming zero —if y has the factor A.

Now putting A=0 in (4) reduces the RHS of (4) to b².  The fifth root of this (V) is B².

So V – A² – B² becomes zero.  This shows that y has the factor A and similarly (by interchangeability) it has the factor B.

Hence y = gAB for some integer g……………………………………………………………………..(7)     

(We also know that g has no higher factor of A and B, because the h term in (4) only has the factors A and B.)       

V⁵, which we get by raising (A² + B²)+ gAB to the power 5 is 

(A² + B²)⁵ + 5gAB(A²+B²)⁴  + 10g²A²B²(A²+B²)³…+ etc……………………………………………..(8)  

Now the “lowest power of AB term” in the expression of (8) is 5gAB(A²+B²) ⁴……………………….(9)

We also know that if we also remove the 2h(a+b) term as well as the A¹⁰ and B¹⁰ terms— from V⁵ in (4) the resulting expression on the RHS of (4) as a plain factor A²B².  [This comes from the 2h² term in (4).]

So the question now reduces to whether the relevant expression in (8) after removing 2h(a+b), i.e.  5gAB(A²+B²)-2h(a+b), has the factor A²B².  This relevant expression may be named ‘W’.

If we divide W by AB it should retain the factor AB.  [‘Should’ here means if (1) is possible.]

W/AB now becomes;   5g(A²+B²) – 10ZV(a+b)………………………………………………………(10)

The key question reduces to whether W/AB has the factor A (or indeed B), which reduces in the A case to whether 5gB² – 10ZVb has the factor A.  [After taking out the explicit factors A.]

Dividing by 5B² —which we know has no factor A— the question becomes: can g -2ZVB³ have the factor A?

At this point we treat V, Z and g as specific unknowns composed of specific combinations of A and B, as before.  Substituting 0 for A will then tell us if the total has a factor A.  If the effect of substituting A=0 is to reduce g -2ZVB³   to zero, this will be a clear sign that g -2ZVB³  has the factor A: if not, it equally clearly does not. Both V and Z will be reduced to their dashed values, V’ and Z’ when A=-0.

Now V = {a² + ab + b² +2h(a+b) +2h²}^1/5. So V’ = B² when A is zero.

Also Z = {(a + b + 2h)/625 }^1/5. So Z’= B/(625) ^1/5  when A is zero.

We know g is an integer, which is all we need to know, because g – 2B³/(625) ^1/5   cannot be zero. [Because the difference between an integer and an irrational number like 1/(625)^1/5 cannot be zero.] Now if (1) is valid, W’ must be zero.  So if we assume (1), this is a contradiction.

SUB-CASE (1 .2)

The second sub-case is when A or B is the unknown with the factor 5. We focus on the case when A has the factor 5, because A and B are virtually interchangeable.

In this case we use result (3’), namely, V = {z² – zb – b² -2h(z+b) +2h²}.

The same line of reasoning as before leads to the result that 

V = Z² – B² + ZBj for some integer j.

So if we take Z¹⁰ and –B¹⁰ and -2h(z+b) away from (Z² – B² + ZBg’)⁵ the resulting expression should have a factor Z²B²… etc.

The reasoning proceeds on the same lines as before and ends up with the result that j – 2B³/(625)^1/5 cannot be zero. This left as a challenge for the reader.  This, too, is a contradiction, assuming (1).

SUB-CASE 3 This is even simpler, because the unknown V as defined by (5) cannot take the factor 5.  We know h has the factor 5, so equation (5) would guarantee that V could have the factor 5, if a² + ab + b² could. However there is no combination of residues for a and b which results in a² + ab + b² having the factor 5.

It follows that the Fermat equation  (a+h)^p + (b+h)^p = (a+b+h)^p   cannot be satisfied when p=5.  The same kind of reasoning also applies in the general case except that for p=7 and larger primes, V can have the factor p.  This means that a different additional line of reasoning is needed. A generalised version of this was given the author’s original 2020 blog in this series. The next blog in this series will offer a simpler version of the same reasoning when p = 7.

NOTICE The above exposition is, in effect, an elementary proof that Fermat’s hypothesis that x^p + y^p = z^p  cannot be satisfied for natural numbers , x, y and z if we confine attention to cases when x+y lacks the factor p. (Simply replace ‘5’ by ‘p’ throughout the argument.)  The case to be considered in the June blog will be the special case when x+y = 0 (mod p).

CHRISTOPHER ORMELL 2^nd May 2022