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Our protagonists are Cartesius (1596-1650) and Fermat (1607-1665). As Judith Grabiner states, in a recommendable text:

“One could claim that, just as the history of Western philosophy has been viewed as a series of footnotes to Plato, so the past 350 years of mathematics can be viewed as a series of footnotes to Descartes’ Geometry.”  (Grabiner) (But remember Michel Onfray‘s observation that followers of Plato have been destroying texts by opponents. (Dutch readers check here.))

Both Cartesius and Fermat were involved in the early development of calculus. Both worked on the algebraic approach without limits. Cartesius developed the method of normals and Fermat the method of adequality.

Fermat and Δf / Δx

Fermat’s method was algebraic itself, but later has been developed into the method of limits anyhow. When asked what the slope of a ray y = s x is at the point x = 0, then the answer y / x = s runs into problems, since we cannot use 0 / 0. The conventional answer is to use limits. This problem is more striking when one considers the special ray that is defined everywhere except at the origin itself. The crux of the problem lies in the notion of slope Δf / Δthat obviously has a problematic division. With set theory we can now define the “dynamic quotient”, so that we can use Δf // Δx = s even when Δx = 0, so that Fermat’s problem is resolved, and his algebraic approach can be maintained. This originated in 2007, see Conquest of the Plane (2011).

Cartesius and Euclid’s notion of tangency

Cartesius followed Euclid’s notion of tangency. Scholars tend to assign this notion to Cartesius as well, since he embedded the approach within his new idea of analytic geometry.

I thank Roy Smith for this eye-opening question:

“Who first defined a tangent to a circle as a line meeting it only once? From googling, it seems commonly believed that Euclid did this, but it seems nowhere in Euclid does he even state this property of a tangent line explicitly. Rather Euclid gives 4 other equivalent properties, that the line does not cross the circle, that it is perpendicular to the radius, that is a limit of secant lines, and that it makes an angle of zero with the circle, the first of which is his definition, the others being in Proposition III.16. I am wondering where the “meets only once” definition got started. I presume once it got going, and people stopped reading Euclid, (which seems to have occurred over 100 years ago), the currently popular definition took over. Perhaps I should consult Legendre or Hadamard? Thank you for any leads.” (Roy Smith, at StackExchange)

In this notion of tangency there is no problematic division, whence there is no urgency to use limits.

The reasoning is:

  • (Circle & Line) A line is tangent to a circle when there is only one common point (or the two intersecting points overlap).
  • (Circle & Curve) A smooth curve is tangent to a circle when the  two intersecting points overlap (but the curve might cross the circle at that point so that the notion of “two points” is even more abstract).
  • (Curve & Line) A curve is tangent to a line when the above two properties hold (but the line might cross the curve, whence we better speak about incline rather than tangent).
Example of line and circle

Consider the line y f[x] = c + s x and the point {a, f[a]}. The line can also be written with c = f[a] – s a:

y – f[a] = s (x a)

The normal has slope –sHwhere we use = -1. The formula for the normal is the line y – f[a] = –sH  (xa). We can choose the center of the circle anywhere on this line. A handy choice is {u, 0}, so that we choose the center on the horizontal axis. (If we looked at a ray and point {0, 0}, then the issue would be similar for {0, c} for nonzero c and thus the approach remains general.) Substituting the point into the normal gives

0 – f[a] = –sH  (ua)

s = (u – a) / f[a]

u + s f[a]

The circle has the formula (x u)² + y² = r². Substituting {a, f[a]} generates the value for the radius r² = (a – (a + s f[a]))² + f[a]² = (1 + s²) f[a]² . The following diagram has {c, s, a} = {0, 2, 3} and thus u = 15 and r = 6√5.

 

descartesMethod of normals

For the method of normals and arbitrary function f[x], Cartesius’s trick is to substitute y = f[x] into the formula for the circle, and then solve for the unknown center of the circle.

(x u)² + (y – 0)² = r²

(x u)² + f[x]² – r² = 0         … (* circle)

This expression is only true for x = a, but we treat it as if it were more general. The key property is:

Since {a, f[a]} satisfies the circle, this equation has a solution for x = a with a double root.

Thus there might be some g such that the root can be isolated:

(x ag [x, u] = 0         … (* roots)

Thus, if we succeed in rewriting the formula for the circle into the form of the formula with the two roots, then we can use information about the structure of the latter to say something about u.

The method works for polynomials, that obviously have roots, but not necessarily for trigonometry and the exponential function.

Algorithm

The algorithm thus is: (1) Substitute f[x] in the formula for the circle. (2) Compare with the expression with the double root. (3) Derive u. (4) Then the line through {a, f[a]} and {u, 0} will give slope –sH. Thus s = (ua) / f[a] gives the slope of the incline (tangent) of the curve. (5) If f[a] = 0, add a constant or choose center {u, v}.

Application to the line itself

Consider the line y f[x] = c + s x again. Let us apply the algorithm. The formula for the circle gives:

(x u)² + (c + s x)² – r² = 0

x² – 2ux + u² + c² + 2csx + s²x² – r² = 0

(1 + s²) x² – 2 (u cs) x +  u² + c² – r² = 0

This is a polynomial. It suffices to choose g [x, u] = 1 + s²  so that the coefficients of are the same. Also the coefficient of must be the same. Thus expanding (xa)²:

(1 + s²) (x² – 2ax +  a²) = 0

– 2 (u cs)  = -2 a (1 +)

u = a (1 +) + cs = a + s (c + sa) = a + s f[a]

which is the same result as above.

A general formula with root x – a

We can deduce a general form that may be useful on occasion. When we substitute the point {af[a]} into the formula for the circle, then we can find r, and actually eliminate it.

(x u)² + f[x]² = r² = (a u)² + f[a

f[x f[a = (a u)² – (x u

(f[x] f[a](f[x] + f[a])  = ((a u) – (x u))  ((a u) + (x u))

(f[x] f[a](f[x] + f[a]) = (a x)   (a + x 2u)

f[x] f[a]  = (a x)  (a + x 2u) / (f[x] + f[a])

f[x] f[a]  = (x a)  (2u – x – a) / (f[x] + f[a])       … (* general)

f[x] f[a]  = (x a) q[x, a, u]

We cannot do much with this, since this is basically only true for x = a and f[x] – f[a] = 0. Yet we have this “branch cut”:

(1)      q[x, a, u] = f[x] – f[a]  / (a x)        if x ≠ a

(2)      q[a, a, u]      potentially found by other means

If it is possible to “simplify” (1) into another expression Simplify[q[x, a, u]] without the division, then the tantalising question becomes whether we can “simply” substitute x = a. Or, if we were to find q[a, a, u] via other means in (2), whether it links up with (1). These are questions of continuity, and those are traditionally studied by means of limits.

Theorem on the slope

We can still use the general formula to state a theorem.

Theorem. If we can eliminate factors without division, then there is an expression q[x, a, u] such that evaluation at x = a gives the slope s of the line, or q[a, a, u] = s, such that at this point both curve and line are touching the same circle.

Proof. Eliminating factors without division in above general formula gives:

q[x, a, u] (2u – x – a) / (f[x] + f[a])

Setting x = a gives:

q[a, a, u] = (u – a) / f[a]

And the above s = (u – a) / f[a] implies that q[a, a, u] = s. QED

This theorem gives us the general form of the incline (tangent).

y[x, a, u] = (x – a) q[a, a, u] + f[a]       …  (* incline)

y[x, a, u] = (x – a) (u – a) / f[a] + f[a

PM. Dynamic division satisfies the condition “without division” in the theorem. For, the term “division” in the theorem concerns the standard notion of static division.

Corollary. Polynomials as the showcase

Polynomials are the showcase. For polynomials p[x], there is the polynomial remainder theorem:

When a polynomial p[x] is divided by (x a) then the remainder is p[a].
(Also, x – a is called a “divisor” of the polynomial if and only if p[a] = 0.)

Using this property we now have a dedicated proof for the particular case of polynomials.

Corollary. For polynomials q[a] = s, with no need for u.

Proof. Now, p[x] – p[a] = 0 implies that – is a root, and then there is a “quotient” polynomial q[x] such that:

p[x] – p[a] = (x a) q[x]

From the general theorem we also have:

p[x] – p[a]  = (x a) q[x, a, u]

Eliminating the common factor (x – a) without division and then setting x = a gives q[a] = q[a, a, u] = s. QED

We now have a sound explanation why this polynomial property gives us the slope of the polynomial at that point. The slope is given by the incline (tangent), and it must also be slope of the polynomial because of the mutual touching of the same circle.

See the earlier discussion about techniques to eliminate factors of polynomials without division. We have seen a new technique here: comparing the coefficients of factors.

Second corollary

Since q[x] is a polynomial too, we can apply the polynomial remainder theorem again, and thus we have q[x] = (x a) w[x] + q[a] for some w[x]. Thus we can write:

p[x] = (x a) q[x] + p[a

p[x] = (x a) ( (x – a) w[x] + q[a] ) + p[a]       … (* Ruffini’s Rule twice)

p[x] = (x a w[x] + (x – a) q[a] + p[a]           … (* Range’s proof)

p[x] = (x a w[x] + y[x, a]                             … (* with incline)

We see two properties:

  • The repeated application of Ruffini’s Rule uses the indicated relation to find both s = q[a] and constant f[a], as we have seen in last discussion.
  • Evaluating f[x] / (x a)² gives the remainder y[x, a], which is the formula for the incline.
Range’s proof method

Michael Range proves q[a] = s as follows (in this article (p406) or book (p32)). Take above (*) and determine the error by substracting the line y = s (x a) + p[a] :

error = p[x] – y = (x a w[x] + (x – a) q[a] – s (x a)

= (x a w[x] + (x – a) (q[a] – s)

The error = 0 has a root x = a with multiplicity greater than one if and only if s = q[a].

Direct application to the incline itself

Now that we have established this theory, there may be no need to refer to the circle explicitly. It can suffice to use the property of the double root. Michael Range (2014) gives the example of the incline (tangent) at x² at {a, a²}. The formula for the incline is:

f[x] – f[a]  = s (x – a)

x² a² – s (x – a) = 0

 (x – a) (x + a s) = 0

There is only a double root or (xa)² when s = 2a.

Working directly on the line allows us to focus on s, and we don’t need to determine q[x] and plug in x = a.

Michael Range (2011) clarifies – with thanks to a referee – that the “point-slope” form of a line was introduced by Gaspard Monge (1746-1818), and that Descartes apparently did not think about this himself and thus neither to plug in y = f [x] here. However, observe that we only can maintain that there must be a double root on this line form too, since {a, f[a]} still lies on a tangent circle.

[Addendum 2017-01-10: The later argument in a subsequent weblog entry becomes: If the function can be factored twice, then there is no need to refer to the circle. But when this would be equivalent to the circle then such a distinction is immaterial.]

Addendum. Example of function crossing a circle

When a circle touches a curve, it still remains possible that the curve crosses the circle. The original idea of two points merging together into an overlapping point then doesn’t apply anymore, since there is only one intersecting point on either side if the circle were smaller or bigger.

An example is the spline function g[x] = {If x < 0 then 4 – x² / 4 else 4 + x² / 4}. This function is C1 continuous at 0, meaning that the sections meet and that the slopes of the two sections are equal at 0, while the second and higher derivatives differ. The circle with center at {0, 0} and radius 4 still fits the point {0, 4}, and the incline is the line y = 4.

descartes-spline

An application of above algorithm would look at the sections separately and paste the results together. Thus this might not be the most useful example of crossing.

In this example there might be no clear two “overlapping” points. However, observe:

  • Lines through {0, 4} might have three points with the curve, so that the incline might be seen as having three overlapping points.
  • Points on the circle can always be seen as the double root solutions for tangency at that point.
Addendum. Discussion

There is still quite a conceptual distance between (i) the story about the two overlapping points on the circle and (ii) the condition of double roots in the error between line and polynomial.

The proof given by Range uses the double root to infer the slope of the incline. This is mathematically fine, but this deduction doesn’t contain a direct concept that identifies q[a] as the slope of an incline (tangent): it might be any line.

We see this distinction between concept and algorithm also in the direct application to Monge’s point-slope formulation of the line. Requiring a double root works, but we can only do so because we know about the theory about the tangent circle.

The combination of circle and line remains the fundamental reason why there are two roots. Thus the more general proof given above, that reasons from the circle and unpacks f[x]² – f[a]² into the conditions for incline and its normal, is conceptually more attractive. I am new to this topic and don’t know whether there are references for this general proof.

Conclusions

(1) We now understand where the double root comes from. See the earlier discussion on polynomials, Ruffini’s rule and the meaning of division (see the section on “method 2”).

(2) There, we referred to polynomial division, with the comment: “Remarkably, the method presumes x ≠ a, and still derives q[a]. I cannot avoid the impression that this method still has a conceptual hole.” However, we now observe that we can compare the values of the coefficients of the powers of x, whence we can avoid also polynomial division.

(3) There, we had a problem that developing p[x] = (x aw[x] + y[x, a] didn’t have a notion of tangency, in terms of Δf / Δx. However, we actually have a much older definition of tangency.

(4) The above states an algorithm and a general theorem with the requirements that must be satisfied.

(5) Cartesius wins from Fermat on this issue of the incline (tangent), and actually also on providing an exact method for polynomials, where Fermat introduced the problem of error.

(6) For trigonometry and exponentials we know that these can be written as power series, and thus the Cartesian method would also apply. However, the power series are based upon derivatives, and this would introduce circularity. However, the method of the dynamic quotient from 2007 still allows an algebraic result. The further development from Fermat into the approach with limits would become relevant for more complex functions.

PM. The earlier discussion referred to Peter Harremoës (2016) and John Suzuki (2005) on this approach. New to me (and the book unread) are: Michael Range (2011), the recommendable Notices, or the book (2015) – review Ruane (2016) – and Shen & Lin (2014).

Cartesius, Portrait by Frans Hals 1648

Cartesius, Portrait by Frans Hals 1648

 

 

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The discussion of Putin’s proof gave me an email from Alexis Tsipras, who just resigned as prime minister of Greece and is busy with the general elections of September 20 soon. Rather than reporting on it, I might as well fully quote it.

To: Thomas
From: Alexis@formerprimeministerofgreece.org
Subject: My proof of Fermat’s Last Theorem
Date: Fri, 28 Aug 2015 11:58:03 +0100
Google Unique Message Identifier: 23DFGA@671

Dear Thomas,

Thank you very much for your discussion of President Putin’s proof when he was a youngster of Fermat’s Last Theorem. I know his mother Vera Putina very well. The Putin family has a vacation home here in Greece, and she can stay there on the condition that she immediately leaves when Putin himself comes down. She has shown me his proof too. I can only agree with your conclusion that it shows how smart President Putin was when he was young.

Putin’s proof inspired me to find a proof too. I am sometimes exhausted by the tough negotiations with the European Heads of State and Government, if not with members of my own party. Thus I resort often to a sanatorium for recuperation. Thinking about such issues like Fermat’s Last Theorem helps to clear my mind from mundane thoughts. I was very happy last Spring to indeed find a much shorter and more elegant proof. 

For the theorem and notation I refer to your weblog. My proof goes as follows.

Theorem. No positive integers n, a, b and can satisfy an + bn = cn for n > 2.

Proof. (Alexis Tsipras, April 31 2015)

Let us assume that an + bn = cn holds, and derive a contradiction.

There are two possibilities: (1) n is even, or (2) n is uneven.

(1) If n is even, then we can write A = an/2 and B = bn/2 and C = cn/2 such that A, B and C are still integers. Then we get the following equation:

A2 + B2 = C2

This equation satisfies the condition that n = 2, and thus it doesn’t satisfy the condition n > 2.

(2) If n is uneven, then we can write A = a(n-1)/2 and B = b(n-1)/2 and C = c(n-1)/2 such that A, B and C are still integers. Then we get the following equation:

a A2 + b B2 = c C2

This equation does not satisfy the form of an + bn = cn so that it falls outside of Fermat’s Last Theorem.

In both cases the conditions of the theorem are no longer satisfied. We thus reject the hypothesis that an + bn = cn holds. Q.E.D.

This is much shorter that President Putin’s proof. And, I prove it while he only came close. I have been hesitating to tell him, fearing that he might become jealous, and be no longer willing to support Greece as he does in these difficult times for my country. Now that you have confirmed how wonderful his proof at only age 12 was, I feel more assured. Will you please publish this proof of mine too, like you did with President Putin’s proof ? I have put my best efforts in this proof, just like at the negotiations with the European Heads of State and Government. Thus I hope that it will be equally convincing, if not more.

After the next elections I will probably be exhausted again. I would like to work on another problem then. Do you have any suggestions ?

Sincerely yours,

Alexis

Fermat and Tsipras (source: wikimedia commons)

Fermat and Tsipras (source: wikimedia commons)

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+)

Fermat (1601 – 1665) and Putin (1952 – ∞) (Source: wikimedia)

Fermat’s Last Theorem, using middle school algebra

Theorem. No positive integers n, a, b and 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 b. Take k = n – 2 > 0. We consider two cases:

(1) a2 + b2c2

(2) a2 + b2 > c2.

(1) When a2 + b2c2 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+2ck+2

ak a2 + bk b2 = ck c2 = ck (a2 + b2 + d)

a2 (akck) + b2 (bkck) = 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 (right rewrites left)

Fermat’s drawing for his proof (RHS re-orders LHS)

Since (1) and (2) cover all possibilities, the theorem holds.

Q.E.D.

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 + bncn 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)2c2, 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[2] = √(a2 + b2) > c. The function has limit f[n → ∞] = a. (See a deduction here.) Thus f[n] is downward sloping from f[2] >  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[2] = √(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.

a (1 + (b/a)^n)^(1/n) and parameter cases

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 (≤ b). Use numerical succession from a2 + b2 > c2.

Given an + bn > cn then prove an+1 + bn+1 > cn+1.

a an + b bnb an + b bn = b (an + bn) > b cnc 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 b2ck c2

ak a2bk a2 + bk a2 + bk b2bk c2ck c2bk c2

a2  (akbk) + bk (a2 + b2c2) ≤ c2 (ck bk)

nonnegative + positive  ≤  nonpositive

Impossible. Thus the Lemma holds for (≤ 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.

Conclusions

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[2] > 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 + bkck) 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 – can + (b cbn < 0

a  an+ b bn  < c (an + bn)

an+1 + bn+1cn+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 (bncn)             (*)

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 <(LHS +) c =(LHS 0)
c > a  (LHS -)
c < (RHS +) (=),
but Case (c b)
impossible opposite signs
c b  (RHS 0) impossible impossible
cn + cn = cn
impossible
c > b  (RHS -) opposite signs impossible (=) 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.