Algebraic Number Theory

After explaining one elementary technique in number theory, I should write about what motivates some of the basic ideas of algebraic number theory by means of a somewhat more complicated proof, namely that 26 is the only integer sandwiched between a square and a cube.

In order to find other numbers similarly sandwiched, we need to solve each of the equations x^2 + 2 = y^3 and x^2 – 2 = y^3. Apart from a few degenerate solutions in which x or y is zero, we only know one integer solution: x = +/-5, y = 3, which corresponds to 25 and 27.

This time, we can’t take quadratic residues, because of that pesky third power. All we can do is tell that x and y are odd; if one is even and one is odd, then the equations say that an odd number and an even number are equal, whereas if they’re both even, then we have a problem since y^3 is divisible by 8, whereas x^2 +/- 2 isn’t even divisible by 4.

It would be great if we could factor the left-hand side… which is a problem, since neither 2 nor -2 is a perfect square. But let’s forget about that hurdle for the moment and try factoring anyway.

We have x^2 + 2 = (x + SQRT(-2))(x – SQRT(-2)). So instead of working just with regular integers – which I’ll call rational integers because they’re all rational numbers – we can work with regular integers, plus the square root of -2. In particular, we work with the set {a + b*SQRT(-2): a and b are integers}, consisting of numbers like 5, 3 + SQRT(-2), -3 – 4SQRT(-2), etc. Since it’s possible to add, subtract, and multiply numbers like this normally, this set forms a ring.

Now, let’s look at the two factors, (x + SQRT(-2)) and (x – SQRT(-2)), a little more closely. In particular, let’s look at any common divisors they have, except the trivial ones 1 and -1. Any common divisor will have to divide their difference, 2SQRT(-2) = -SQRT(-2)^3. So this common divisor is SQRT(-2), 2, or 2SQRT(-2), which is divisible by SQRT(-2).

That means that x + SQRT(-2) is divisible by SQRT(-2), or, if you will, that x is divisible by SQRT(-2). But x/SQRT(-2) = (x/2)SQRT(-2), and we’ve already proven that x is odd, so there’s a contradiction, and the two factors have no common divisors.

If they have no common divisors, then they’re both cubes. This is fairly common sensical: any prime factor that divides the first factor has to divide y^3. So its cube must divide y^3, too, which means it divides the first factor, or else the first and second factor are both divisible by that prime.

So there’s a number, call it a + bSQRT(-2), such that (a + bSQRT(-2))^3 = x + SQRT(-2). Expanding the left-hand side, we get that a^3 + 3a^2*bSQRT(-2) – 6ab^2 – 2b^3*SQRT(-2) = x + SQRT(-2). Both the rational-integer and the SQRT(-2) parts must be equal, so we have 3a^2*b – 2b^3 = 1, where a and b are rational integers.

Now we have enough to apply simpler tricks. The left-hand side is divisible by b, so b has to be +/-1. If it’s -1, then we get -3a^2 + 2 = 1, or 3a^2 = 1, which is absurd since a is a rational integer. If b = 1, then we have 3a^2 – 2 = 1, or 3a^2 = 3, which means a = +/-1.

If a = 1, then (a + SQRT(-2))^3 = -5 + SQRT(-2), so x = 5. Similarly, if a = -1, then x = -5. Then y = 3 and we get 26.

We can do exactly the same thing with the other equation, only this time we work with SQRT(2). All the steps work exactly the same, only we end up with 3a^2*b + 2b^3 = 1. In that case, b = 1 gives 3a^2 = -1, a contradiction, and b = -1 gives 3a^2 = -3, another contradiction.

So 26 is really the only number sandwiched between a square and a cube… supposedly. I say “supposedly” because I lied to you a bit – actually, there’s one or two very important things left to check that I didn’t check here. In this case they work, but they don’t have to, and I need to show that they work. But that’s for next time.

7 Responses to Algebraic Number Theory

  1. Bryan says:

    Pardon my ignorance, but I know next to nothing about serious math. What is the significance/relevance of this sort of abstract number theory? Is it “math for math’s sake” or is there some sort of advanced application?

  2. Alon Levy says:

    This particular tidbit is entirely recreational. But deeper aspects of the theory, such as a lot of the mathematics that’s relevant to the proof of Fermat’s Last Theorem (an entirely useless result not only in the real world but also in other areas of mathematics), have cryptographic applications. Elliptic curve encryption is public-key, like RSA, but whereas RSA is weaker by countless orders of magnitude than private-key encryption with the same key length, elliptic curve isn’t.

    Also, some of the more computational aspects of abstract algebra are useful in coding theory.

  3. […] I promised that I’d explain later the crucial complication with the proof that 26 is the only integer sandwiched between a square and a cube. In fact, that complication permeates the entire discipline, and the more advanced methods it uses seem to largely be about getting around it. […]

  4. Mathew says:

    Very cute ;). I googled the proof while I was reading Simon Singh’s Fermat’s Last Theorem for the third time.

  5. Joey says:

    Hey great work on the proof. Really helped me a lot. Thank you.

  6. Ashutosh says:

    I too looked up this proof after reading Simon Singh’s book! Thanks a lot for the proof. Would be great if you could please explain to me the further complication that permeates the logic. Maybe, it would require advanced mathematics to go into detail but can you at least describe in principle, the complication that was overlooked in this proof.

    I have a hunch that it is to do with the fact that complex numbers do not obey the unique factorization principle i.e the fact that there is a unique way of factorizing a number into prime factors. I guess the above proof dabbled with trying to factorize into complex factors and hence may be faced with proving the rigor of this process. Correct me if I am wrong!

  7. JoseBrox says:

    Nice entry! Allow me to make a little clarification:

    You say in the introduction that there is just one solution “apart from a few degenerate solutions in which x or y is zero”.

    That is not exactly true: in fact there are no solutions with x=0 or y=0! I think that what you meant is that the only other solution is x=1, y=-1, which is quite trivial and sees the number 0 sandwiched (x=-1=y is a solution to the second equation, but not to the original problem).

    This solution did not appear in your exposition because it arises from the consideration of the cases in which y has no prime factors, i.e., when y=1 or y=-1.

    Jose Brox

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