PrepShorts · Study sheet · Class 8 Mathematics · Chapter 2, The Baudhāyana-Pythagoras Theorem
Chapter 2 · The Baudhāyana-Pythagoras Theorem
Fermat's Last Theorem: the same question one power up
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Squares split into two smaller squares over and over. Try it one power up and nobody found a single example in 350 years.
The idea
§2.5 has just shown that a square splits into two squares in infinitely many ways. The obvious next question is whether a cube splits into two cubes, and the answer turns out to be the exact opposite: never, and never for any power above two. So the abundance of Baudhāyana triples is not a general fact about powers at all — it is a peculiarity of the exponent 2, and the boundary sits precisely where this chapter has been working. The other half of the topic is what it took to know that. No search can establish a "never", however far it runs, which is why a statement anybody can understand went unproved for more than three centuries.
What you should be able to do
- State the equation xⁿ + yⁿ = zⁿ and the conditions the chapter attaches to it: natural numbers, and an exponent above 2
- Explain how the question arose from the study of integer triples
- Contrast the case n = 2, with infinitely many solutions, against every case above it, with none
- Explain why checking cubes one after another can never settle the question, however many are checked
- Recount the chapter's account of Fermat's marginal note and what became of the proof it claimed
- Give the chapter's timeline: Fermat in the 17th century, over three centuries of failure, Wiles reading about the problem in 1963, and his proof in 1994
- Distinguish "no proof was found" from "no proof existed", and say which the chapter actually asserts
- Say why this section sits at the end of a chapter about Baudhāyana
Words to know
| Term | Definition in one line | First introduced |
|---|---|---|
| Fermat's Last Theorem | the statement that xⁿ + yⁿ = zⁿ has no natural-number solution once n is above 2 | printed in bold in this chapter (Part II, §2.6, p.51) |
| natural numbers | the counting numbers the equation's three unknowns must be | printed in this chapter (Part II, §2.6, p.51) |
| perfect cube | a number that is some whole number multiplied by itself three times | printed in this chapter (Part II, §2.6, p.51) |
| fourth power | a number that is some whole number raised to the fourth | printed in this chapter (Part II, §2.6, p.51) |
| open problem | a question posed but not yet settled — the chapter's section title calls this one long-standing | printed as the section heading in this chapter (Part II, §2.6, p.50) |
| Elements | Euclid's book, cited earlier in this chapter for the proof about √2 | printed earlier in this chapter (Part II, §2.3, p.39) |
| The Last Problem | the book by Eric Bell that the chapter says Wiles read as a child | printed in this chapter (Part II, §2.6, p.51) |
| exponent | the raised number saying how many times the base is multiplied by itself | not printed anywhere in this chapter, which writes of powers and of n and never names the raised position |
| counterexample | a single case that would demolish the statement, if one existed | an added term; not printed in this chapter |
| conjecture | a general claim advanced before proof — what this was for 300 years | printed earlier in this chapter (Part II, §2.5, p.48) |
Where people slip up
- "It was called a theorem, so Fermat proved it." The chapter is careful: the claimed proof was never found, and it was proved by someone else 300-odd years later. The name is historical. An explanation must not let "theorem" imply Fermat had one, and should not assert that he did not either — the chapter takes no position and neither should the explanation.
- "Nobody could find a solution, so there isn't one." Backwards, and it is the central logical point of the section. Not finding is not the same as not existing. This is why the near-miss 728 and 729 belongs in the explanation: it shows how close a search can come while proving nothing.
- "So mathematicians spent 300 years checking numbers." They did not, or at least that is not what took the time. Checking cases could never have finished the job. What was needed was an argument covering every exponent at once, which is why the problem was hard and why the eventual proof uses mathematics far beyond this chapter.
- "A ten-year-old proved Fermat's Last Theorem." He was ten when he read about it and an adult mathematician when he proved it, three decades later. The chapter says exactly this, and an explanation that compresses it turns a good story about persistence into a false one about precocity.
- "The theorem says powers never add up." It says nothing of the kind. Powers add up constantly; what fails is the demand that the total be the same power of a whole number. And for n = 1 and n = 2 even that succeeds endlessly. Section 10 exists to stop this.
- "There might still be a huge counterexample." Not any more; the statement is proved. Before 1994 that was a live possibility and it is part of why the problem resisted — the numbers involved could have been enormous.
- "This is examinable content." Treat it as context and culture rather than as technique. There is nothing here to compute. Its value is that it shows a student what an open problem is, and that the arithmetic they have just done sits at the edge of something unresolved for centuries.
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Worked answers to this chapter’s exercises
Transcript1,342 words
A square splits into two smaller squares, in whole numbers, over and over. Nine plus sixteen is twenty-five. Twenty-five plus a hundred and forty-four is a hundred and sixty-nine. Search every pair up to two hundred and you find a hundred and twenty-seven of them, and the supply never runs out. That is a lot of solutions for one equation, and it is easy to walk away thinking that is just what powers do.
So let us test that thought, by changing exactly one thing. Not the shapes. Not the numbers. Only the power. Instead of a square splitting into two squares, ask whether a cube splits into two cubes. Here are the first ten cubes. One, eight, twenty-seven, sixty-four, a hundred and twenty-five, two hundred and sixteen, three hundred and forty-three, five hundred and twelve, seven hundred and twenty-nine, and a thousand. The question is exactly the one we just answered for squares. Pick two from that list. Add them. Does the total appear on the list as well?
For squares, that happened constantly. Let us find out what happens here. One plus eight is nine. Not a cube. Eight plus twenty-seven is thirty-five. Not a cube. Twenty-seven plus sixty-four is ninety-one. Sixty-four plus a hundred and twenty-five is a hundred and eighty-nine. Neither is a cube. Try skipping about instead of taking neighbours. One plus twenty-seven is twenty-eight. One plus sixty-four is sixty-five. Eight plus sixty-four is seventy-two.
Nothing. Not one of them lands. Go higher, where the cubes are far apart and there is more room. A hundred and twenty-five plus two hundred and sixteen is three hundred and forty-one. Five hundred and twelve plus seven hundred and twenty-nine is twelve hundred and forty-one. Still nothing. Ten pairs tried, and every single total falls in a gap. Except for one, which is worth stopping on. Two hundred and sixteen is six cubed. Five hundred and twelve is eight cubed. Add them and you get seven hundred and twenty-eight.
Seven hundred and twenty-eight. And nine cubed is seven hundred and twenty-nine. It misses by one. Out of numbers in the hundreds, the total lands one single unit away from a cube. That feels like an argument. It is not one. Seven hundred and twenty-eight is not a cube, and one away is not on the list. Remember how close this came, though. We are going to need it. Ten pairs is nothing. Let us search properly.
Take every pair of whole numbers up to two hundred. That is twenty thousand one hundred pairs. Cube both, add them, and check whether the total is a cube. Not one of the twenty thousand one hundred works. Now, that could mean two very different things. It could mean there is nothing there. Or it could mean the search itself is broken. So run the same search, unchanged, one power down, on squares.
It comes back with a hundred and twenty-seven. The search is not broken. When there is something to find, it finds it. Twenty thousand pairs, a search we have just seen work, and nothing at all. Here is the trap, and almost everybody walks into it. A search reports what it did not meet. It never reports what is not there. Watch how badly that can go. Take a number, square it, add the number again, and add forty-one.
At nothing, that is forty-one, which is prime. At one, forty-three, prime. At two, forty-seven. Then fifty-three, sixty-one, seventy-one. All prime. Keep going and they stay prime. Ten in a row. Twenty. Thirty. Forty in a row, every one of them prime, with no exception anywhere. At forty, it gives sixteen hundred and eighty-one. And sixteen hundred and eighty-one is forty-one times forty-one. Forty confirmations in a row, and the forty-first case knocked the whole thing over.
It is not a one-off. Take two to a power, and subtract one. Two cubed less one is seven, prime. Two to the fifth, less one, is thirty-one, prime. Two to the seventh, less one, is a hundred and twenty-seven, prime. Two to the eleventh, less one, is two thousand and forty-seven. Which is twenty-three times eighty-nine. So take six statements, search each one to a small limit, then search it again to a larger one.
Read what the small search reports as if it were the truth, and on four of the six you are right. On two of them you are wrong. The two we just watched break. Four out of six is not a good enough record to settle a never. And twenty thousand pairs are not different in kind from forty. So write the question down properly. Some number to a power, plus another number to that same power, equals a third number to that same power.
All three whole and positive, and the power bigger than two. Be careful what that is claiming, because it is not that powers never add up. Powers add up all the time. Twenty-seven plus sixty-four plus a hundred and twenty-five is two hundred and sixteen. Three cubes making a cube exactly. One plus seventeen hundred and twenty-eight is seventeen hundred and twenty-nine. So is seven hundred and twenty-nine plus a thousand. The same total, from two different pairs of cubes.
That total is still not a cube. It misses twelve cubed by one, just like before. What fails is narrower than it looks. Two of them, of the same power, adding to that same power of a whole number. In the sixteen hundreds, a French lawyer named Pierre de Fermat wrote a note in a margin. He said no such numbers exist, for any power above two, and that he had a proof.
He did not write the proof down. It has never been found, and there is no way to know whether he had one. The claim went out into the world anyway, and people started trying. He wrote it somewhere in that century, and it was settled in nineteen ninety-four. So whichever year it was, somebody waited at least two hundred and ninety-four years, and possibly three hundred and ninety-three. Every one of those years, the answer was already whatever it was. Nobody could show which.
In nineteen sixty-three, a boy of ten read about this problem. He could understand the question completely. Anyone here can. That is the strange thing about it. He decided he would be the one to settle it. He proved it in nineteen ninety-four. His name is Andrew Wiles. That is thirty-one years later. He was ten when he read about it and forty-one when he finished. And he did not finish it by checking numbers. Centuries of checking numbers had found nothing, and finding nothing was never going to be enough.
He finished it with an argument. So the answer is no, everywhere, for every power above two, and there is no enormous counterexample waiting somewhere out of reach. Now put the three cases side by side, because this is the part worth carrying away. Run one search, to one limit, at one power after another. At the first power it is just addition. Two plus three is five, and every pair works. Up to two hundred, ten thousand solutions.
At the second power, a hundred and twenty-seven, and endlessly many beyond. At the third power, nothing. At the fourth, nothing. Fifth, nothing. Sixth, nothing. Unlimited, unlimited, then nothing at all. And the step down happens exactly between two and three. So all those whole-number triples were never a fact about powers. They were a fact about the number two, and about nothing else. One power up, the same question, asked the same way, and the answer changes from endlessly many to none.
The boundary sits precisely where we have been working the whole time. And notice what could never have told you that. Not the ten pairs. Not the twenty thousand. Not the near miss at seven hundred and twenty-eight. Searching harder was always the wrong instrument. Somebody had to stop looking and start arguing.
Where this fits
Taken from the notes each video was made from, not from the reading order — these are the ideas this one rests on and the ones that later rest on it.
Builds on
- Baudhāyana-Pythagoras triples, and how to make infinitely manyClass 8 · Ch 2, The Baudhāyana-Pythagoras Theorem
Either side of this one
- Applying the theorem: the lotus-in-the-lake problem from the LīlāvatīClass 8 · Ch 2, The Baudhāyana-Pythagoras Theorem