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On Mathworld it says some diophantine equations can be solved using algebraic numbers. I know one example which is factor $x^2 + y^2$ in $\mathbb{Z}[\sqrt{-1}]$ to find the Pythagorean triples.

I would be very interested in finding some examples of harder equations (not quadratic) which are also solved easily using algebraic numbers. Thank you!

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I haven't seen Section 7 of Lenstra's article mentioned: – Samuel Hambleton Oct 13 '11 at 11:31
up vote 3 down vote accepted

Euler famously used algebraic integers to solve diophantine problems, though not always exactly right.

For example, in article 182 of Chapter XII, Part II, of his Elements of Algebra (p. 396 in my brand new Cambridge Library Collection edition), he writes:

Let, therefore, the formula $x^2+cy^2$ be proposed, and let it be required to make it a square. As it is composed of the factors $(x+y\sqrt{-c})(x-y\sqrt{-c})$, these factors must either be squares, or squares multiplied by the same number. For, if the product of two numbers, for example, $pq$, must be a square, we must have $p=r^2$ and $q=s^2$; that is to say, each factor is of itself a square; or $p=mr^2$ and $q=ms^2$; and therefore these factors are squares multiplied by the same number. For which reason, let us make $x+y\sqrt{-c} = m(p+q\sqrt{-c})^2$; it will follow that $x-y\sqrt{-c} = m(p-q\sqrt{-c})^2$, and we shall have $x^2+cy^2 = m^2(p^2+cq^2)^2$, which is a square. Farther, in order to determine $x$ and $y$, we have the equations $x+y\sqrt{-c} = mp^2 + 2mpq\sqrt{-c} -mcq^2$, and $x-y\sqrt{-c} = mp^2-2mpq\sqrt{-c} - mcq^2$; in which $x$ is necessarily equal to the rational part, and $y\sqrt{-c}$ to the irrational part; so that $x=mp^2 - mcq^2$ and $y\sqrt{-c} = 2mpq\sqrt{-c}$, or $y=2mpq$; and these are the values of $x$ and $y$ that will transform the expression $x^2+cy^2$ into a square, $m^2(p^2+cq^2)^2$, the root of which is $mp^2+mcq^2$.

(Note that Euler in fact finds sufficient conditions, but not necessary ones, here, and there is an implicit assumption of unique factorization). In Article 187, he proceeds along the same lines to discuss when $ax^2+cy^2$ will be a cube, working with the factorization $ax^2+cy^2 = (x\sqrt{a}+y\sqrt{-c})(x\sqrt{a} - y\sqrt{-c})$.

Euler himself knew this did not quite work out in general: in Article 195 he discusses when $2x^2-5$ is a cube; his method yields the answer that it is never a cube, but yet he notes that $x=4$ gives $2x^2 - 5 = 27 = 3^3$; his "investigation" into the failure involves some rather strenuous hand-waving, concluding that the problem is that we have a difference instead of a sum as before.

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+1 for "rather strenuous hand-waving." Great imagery. – Qiaochu Yuan Jan 19 '11 at 20:16

Perhaps the simplest example is the parametrization of primitive Pythagorean triples $\rm x^2 + y^2 = z^2\:.$ The essence of the proof is: $\rm\ x+y\ i,\ \ x-y\ i\ $ are coprime factors of a square in a UFD, so they must themselves be squares (up to unit factors). So $\rm\ x + y\ i\ =\ (m + n\ i)^2 =\ m^2 - n^2 + 2\: m\: n\ i $. Similarly one can solve low degree cases of Fermat's Last Theorem by employing analogous factorizations over certain rings of algebraic integers. For example, Gauss showed there are no solutions for exponent 3 by working in the ring of integers of $\rm\ \mathbb Q(\sqrt{-3})\:,\: $ and Dirichlet did similarly for exponent 5 using $\rm\ \mathbb Q(\sqrt{5})\:.$ Later Kummer generalized these techniques to handle all regular prime exponents by working over rings of cyclotomic integers. For a nice exposition see Ribenboim: 13 lectures on Fermat's last theorem. Weil nicely summarizes the essence of these techniques in his Number Theory, Ch.IV,S.VI,p.335:

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For the integer case of coprime divisors of powers, see this answer for a proof by induction, and see this answer for a proof via gcds. – Bill Dubuque Feb 13 '15 at 3:11

The classic example is certain Mordell curves. I can never quite remember which ones have this property off the top of my head, but for example I think $y^2 = x^3 - 1$ is one of them. (Write it as $y^2 + 1 = x^3$ and factor the LHS in $\mathbb{Z}[i]$.) There might be a list in some books on Diophantine equations and/or algebraic number theory and/or elliptic curves.

There is also the regular case of Fermat's last theorem. I don't know if this counts as "easy," though.

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For some worked examples of Mordell's equation, using algebraic numbers, see Section 3 of – KCd Feb 1 '11 at 5:44

Fermat's Last Theorem was originally "proved" by using some elementary arguments and the "fact" that the ring of integers of $\mathbb{Q}(\zeta_p)$ is a UFD (which it is not), where $\zeta_p$ is a primitive $p$th root of unity. These arguments do hold when $\mathcal{O}_p$ is a UFD though; for a thorough account of the argument, see Chapter 1 of Marcus's Number Fields.

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An interesting example which appears in section $4.9$ of Stewart and Tall's Algebraic Number Theory and Fermat's Last Theorem is the solution of the Ramanujan-Nagell equation given by

$$x^2 + 7 = 2^n$$

The idea is to work over the quadratic field $\mathbb{Q}(\sqrt{-7})$ and use the fact that it has unique factorization in order to compare two factorizations at the core of the argument.

The argument needs to take care of different cases separately and does not run as "smoothly" as the arguments in some of the examples mentioned in the other answers, but that's probably in some sense due to the fact that this equation involves an exponential, which only makes the already difficult problem of solving a Diophantine equation worse.

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There are many examples, some of which I see you have already gotten :-)

Here is one example I had come across recently:

Show that the following diophantine equation, for integral $\displaystyle a,b$, has no non-trivial solutions.

$$3b^4 + 3b^2 + 1 = a^2$$

I came across this while trying to find an "elementary" solution to the diophantine equation

$$y^2 = x^3 - 1$$

which is easily solved using $\displaystyle \mathbb{Z}[i]$, see this here: Integral solutions to $y^{2}=x^{3}-1$.

Of course, there might be an easy elementary proof for $\displaystyle 3b^4 + 3b^2 + 1 = a^2$, but, in fact this is closely related to the Pell's equation $\displaystyle 3x^2 + 1 = y^2$ and amounts to showing that the non-trivial odd values of $\displaystyle x$ are such that $\displaystyle \frac{x-1}{2}$ cannot be a perfect square. This itself can be cast into a problem of the terms of a linear recurrence never being perfect squares. So I do expect this problem to give some resistance to an elementary solution.

To solve the problem:

Mutiply by $\displaystyle 9b^2$

$$3b^2(9b^4 + 9b^2 + 3) = (3ab)^2$$

put $\displaystyle x = 3b^2 + 1$ and $\displaystyle y = 3ab$

$$(x-1)(x^2 +x + 1) = y^2$$

$$x^3 - 1 = y^2$$

which we can show has no non-trivial solutions easily, by using $\displaystyle \mathbb{Z}[i]$.

Note: If one can prove non-existence of $\displaystyle 3b^4 + 3b^2 + 1 = a^2$ by other means, we will have found a different proof of non-existence of $\displaystyle x^3 - 1= y^2$, because the existence of non-trivial solutions to $\displaystyle x^3 - 1= y^2$ implies the existence of non-trivial solutions to $\displaystyle 3b^4 + 3b^2 + 1 = a^2$. (That is how I came across this).

I will leave that to you.

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Does this mean an elementary treatment of $y^2 = x^3-1$ should be called a Moronic approach? – KCd Feb 1 '11 at 5:48
@KCd: Only if it works :-P – Aryabhata Feb 1 '11 at 6:28

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