The door rang. I was surprised to see Vera Putina. It appeared that Putin’s mother was visiting her granddaughter’s fiancé’s family in Holland. “It is not safe for me to go to Moscow,” she explained, expressing the sentiment of many.
When she was settled in the safety in my living room with a good cup of Darjeeling, it also appeared that there was more.
VP: “I am upset. The Western media depict my son only as a sportsman. They show him doing judo, riding horses, fighting bears, and the last week they featured him as a diver in a submarine. Of course he is very athletic, but he is also a smart man. I want you to look at his intellectual side too.”
Me: “It is fine of you to ask, because there indeed is a general lack of awareness about that.”
VP: “Let me tell you ! When my father had his love affair with Grand Duchess Kira Kirillovna of Russia [see the Putin family tree here], one of their secret meeting places was in the royal archives of the Romanovs. Sometimes my father, because he was unemployed, took some of the old documents to sell on the market. You know, my mother Kira had an expensive taste.”
VP said this without blinking an eye. The unperturbedness when taking other people’s possessions and territories must have a family origin.
VP: “My father once found the application letter by Pierre de Fermat for membership of the St. Petersburg mathematical society. It detailed his proof of his Last Theorem.”
Me: “Ah, that might explain why he never published it ! He used it for his application, and this got lost in the archives ?!”
VP: “I don’t know about that. My father sent it in for the Wolfskehl Prize of 100,000 gold marks, but it was rejected for it didn’t satisfy the criterion of having been published in a peer-reviewed journal.”
Me: “This is historically very interesting. If you still have that letter by Fermat, no doubt a historical journal will gladly publish it. It doesn’t matter on content since Andrew Wiles now proved it.”
VP: “No, no, no !” She gestured with passion as an Russian woman can do. “Fermat is not important ! It is what Vladimir did ! When he was twelve, he also looked at Fermat’s letter, and he found an omission ! Moreover, he worked on the problem himself, and almost solved it. Here, I brought along the papers to prove it to you.”
She delved into the bag that she had brought along and produced a stack of papers. I also saw a wire bound notebook such as children use in school.
Me: “Almost solving means not solving. Mathematics is rather strict on this, gospodina Putina. But it is historically interesting that you have Fermat’s original proof and that your son worked on it.”
VP: “For this, he had to learn Latin too !’
She gave me the stack. There was a great deal of difference between her nonchalant and triumphant handing over of the papers and my hesitant and rather reverent accepting of them.
VP: “You look it over, and inform the Western media that my son almost solved Fermat’s Last Theorem when he was only twelve ! If I hadn’t told him that he had to go to his judo lessons, he would have finished it for sure !”
She said the latter as proof that she had been a good mother, but also with a touch of regret.
Confronted with such motherly compassion I could only respond that I would oblige. Hence, below is Vladimir Putin’s proof. First I translate Fermat’s own proof from Latin (also using the Russian transcript that Putin made) and then give Putin’s correction.
Fermat (1601 – 1665) and Putin (1952 – ∞) (Source: wikimedia)
Fermat’s Last Theorem, using middle school algebra
Theorem. No positive integers n, a, b and c can satisfy the equation an + bn = cn for n > 2.
Proof. (Pierre de Fermat, April 31 1640, letter to czar Michael I of Russia)
Without loss of generality a ≥ b. Take k = n – 2 > 0. We consider two cases:
(1) a2 + b2 ≤ c2
(2) a2 + b2 > c2.
(1) When a2 + b2 ≤ c2 then a2 + b2 + d = c2 for d ≥ 0
Then a < c and b < c. Then also ak < ck and bk < ck for k > 0.
If the theorem doesn’t hold, then there is a k > 0 such that:
ak+2 + bk+2 = ck+2
ak a2 + bk b2 = ck c2 = ck (a2 + b2 + d)
a2 (ak – ck) + b2 (bk – ck) = d ck ≥ 0
negative + negative ≥ 0
Impossible. Thus the theorem holds for (1).
(2) If a2 + b2 > c2 then obviously (see the diagram) for higher powers too: an + bn > cn.
Fermat’s drawing for his proof (RHS re-orders LHS)
Since (1) and (2) cover all possibilities, the theorem holds.
Putin’s correction, age 12
The comment by schoolboy Vlad on this proof is:
“While (2) is obvious, you cannot rely on diagrams, and you need to fully develop it. At least I must do so, since I find the diagram not so informative. I also have problems reading maps, and seeing where the borders of countries are.”
Hence, young Putin proceeds by developing the missing lemma for (2).
Lemma. For positive integers n, a, b and c: if a2 + b2 > c2 then an + bn ≠ cn for n > 2.
Proof. (Vladimir Putin, October 7 1964)
Without loss of generality a ≥ b. Take k = n – 2 > 0.
If a2 + b2 > c2 then a + b > c. (Assume the contrary: a + b ≤ c then a2 + b2 < (a + b)2 ≤ c2, which contradicts a2 + b2 > c2.)
Expression an + bn > cn is equivalent to (an + bn)1/n > c. The LHS can be written as:
f[n] = a (1 + (b / a)n)1/n with a ≥ b.
This Lemma has the Pythagorean value f = √(a2 + b2) > c. The function has limit f[n → ∞] = a. (See a deduction here.) Thus f[n] is downward sloping from f > a to limit value a. We have two cases, drawn in the diagram below.
Case (A) Diagram LHS: c ≤ a, so that there will never be an intersection f[n] = c.
Case (B) Diagram RHS: a < c < f = √(a2 + b2). There can be an intersection f[n] = c, but possibly not at an integer value of n. Observe that this case also provides a counterexample to Fermat’s claim that “obviously” f[n] > c, for, after the intersection f[n] < c. Young Putin already corrects the great French mathematician ! This is a magnificent result of the future President of the Russian Federation, at such a young age. His grandfather’s Marinus van der Lubbe’s submission to the Wolfskehl Prize would also have failed on this account.
f[n] = a (1 + (b / a)^n)^(1/n) and parameter cases
At this point, young Putin declares that Case (A) on the LHS is proven, based upon above considerations. He adds:
“I accept this proof on the LHS, even though I have difficulty understanding that limits or borders should not be transgressed.”
As so often happens with people who are not entirely sure of their case, the schoolboy then develops the following simple case, just to make certain.
Case (c ≤ b). Use numerical succession from a2 + b2 > c2.
Given an + bn > cn then prove an+1 + bn+1 > cn+1.
a an + b bn ≥ b an + b bn = b (an + bn) > b cn ≥ c cn
Thus the Lemma holds for this case.
To be really, really, sure, Putin adds an alternative proof that assumes the contrary:
ak a2 + bk b2 ≤ ck c2
ak a2 – bk a2 + bk a2 + bk b2 – bk c2 ≤ ck c2 – bk c2
a2 (ak – bk) + bk (a2 + b2 – c2) ≤ c2 (ck – bk)
nonnegative + positive ≤ nonpositive
Impossible. Thus the Lemma holds for (c ≤ b).
It would have been better when he had looked at Case (B) on the RHS, notably by proving that f[n] = c cannot hold for only integers.
At this point in his notebook, young Putin writes:
“I have to go to judo training. Perhaps I will continue tomorrow.”
I have looked in the remainder of the notebook but did not find further deductions on Fermat. Apparently the next day young Putin continued with what was more on his mind. It appears that he had a fantasy land called Dominatia in which he played absolute master, and it took much of his time to determine what was happening there. Something of the unruly nature of the natural numbers however must have stuck in his mind. In a perfect fantasy land everything is already as wished, but in young Putin’s Dominatia land he fantasizes about unruly citizens who must be put under control.
The above supports the following conclusions:
- The theorem & lemma are not yet proven for Case (B) on the RHS. We must still rely on Andrew Wiles.
- Nevertheless, Vladimir Putin doesn’t do just sports but also has amazing intellectual powers, at least when he was at age twelve.
- Fermat’s original own proof of his theorem seems to have had a serious error, but it is not precluded that it was only chance that it did not get published (with or without corrections).
- Fermat’s Last Theorem has dubious value for education. It seems more important to develop the notion of limits, and in particular the notion that you should not transgress borders. When students do not understand this properly at a younger age then this may cause problems later on.
Appendix 1. Case (c > a ≥ b)
It may be nice to see how f[n] = c is sandwiched, when a + b > f > c > a ≥ b.
Case (c > a ≥ b) There is a point f[n] = c or an + bn = cn for reals but perhaps not for integers.
(i) At the intersection:
ak a2 + bk b2 = ck c2
Take ak c2 + bk c2 and substract the above on both sides:
(c2 – a2) ak + (c2 – b2) bk = (ak + bk – ck) c2
positive + positive = ?
The latter must be positive too, and hence: ak + bk > ck
Thus, assuming that the theorem doesn’t hold for n requires that it holds for k = n – 2.
(ii) After the intersection: Since f[n] is downward sloping we have f[n+1] < f[n] = c. Reworking gives:
an+1 + bn+1 < cn+1
Another way to show this is:
(a – c) an + (b – c) bn < 0
a an+ b bn < c (an + bn)
an+1 + bn+1 < cn+1
Comparing (i) and (ii) we see the switch from > to <.
Appendix 2. Parameter restrictions in general
Assume that an + bn = cn holds. There are restrictions for this to occur, notably by the remarkable product:
(an – bn) (an + bn) = (an – bn) cn
a2n – b2n = (an – bn) cn
an (an – cn) = bn (bn – cn) (*)
For example: when c = a, then an + bn = cn is only possible in (*) if c = a = b, but this is actually also impossible because it requires that cn + cn = cn. The table collects the findings, with the LHS and RHS now referring to equation (*).
|an + bn = cn
||c < a (LHS +)
||c = a (LHS 0)
|c > a (LHS -)
|c < b (RHS +)
but Case (c ≤ b)
|c = b (RHS 0)
cn + cn = cn
|c > b (RHS -)
||(=) the only risk
This table actually also proves Case (A) that Putin took for granted. Only Case (B) remains, and requires proof that f[n] = c cannot hold for only integers.