Show that infinitely many integer solutions $(a, b)$ exist for:


Is there a contrary proof against this? Thanks for help.

  • $\begingroup$ Note 2a^2 is not a square. $\endgroup$ – Jacob Wakem Mar 25 '17 at 1:37

Just to provide an algebraic viewpoint on this problem;

If we rearrange $2a^2 + 1 = b^2$ we have

$$b^2 - 2a^2 = 1$$

or, equivalently

$$(b+a\sqrt{2})(b-a\sqrt{2}) = 1.$$

These are elements of the number field $\mathbb{Q}[\sqrt{2}] = \left\lbrace p + q\sqrt{2}: p, q \in \mathbb{Q}\right\rbrace$, but since we only care about the integer values for which your equation holds, we look to the ring of integers $\mathcal{O}_{\mathbb{Q}[\sqrt{2}]}$. Since $2 \not\equiv 1 \mod 4$, this ring is simply $\mathbb{Z}[\sqrt{2}] = \left\lbrace k + l\sqrt{2}: k,l \in \mathbb{Z}\right\rbrace$. For an element $\alpha = b + a\sqrt{2}$ define the function $N_{\mathbb{Z}[\sqrt{2}]}(\alpha) = (b + a\sqrt{2})(b - a\sqrt{2})$, called the norm of an element. It can be shown that $N(\alpha) = 1 \iff \alpha $ is a unit.

Hence, the problem translates to showing that the group of units $\mathbb{Z}[\sqrt{2}]^\times$ of $\mathbb{Z}[\sqrt{2}]$ has infinite order.

The details of these calculations are "fleshed out" here.

  • $\begingroup$ Just an aside, the abstract in the notes that I linked give this description of Dirichlet's Unit Theorem "Dirichlect’s Unit Theorem establishes the structure of the units of a number field as a finite abelian group." This is not correct (maybe a typo), instead it should read "Dirichlect’s Unit Theorem establishes the structure of the units of a number field as a finitely generated abelian group." $\endgroup$ – Edward Evans Mar 25 '17 at 21:54

This is the Pell's equation with $n=2$ i.e. $b^2-2a^2 = 1$.

You can check that $a=2, b=3$ is the smallest solution of this problem. This is called the fundamental solution, and in fact all solutions of the Pell's equation can be generated by this solution.

You can look up Wikipedia for more details, but as it turns out, there are infinitely many $a_k,b_k$ for which this happens, and the relationship in this case is: $$ b_k + a_k\sqrt 2 = (3+2\sqrt 2)^k $$

You can check that if we expand the RHS by binomial theorem and compare coefficients, we get the required quantities.

Example: If we take $k=2$,then $(3+2\sqrt2)^2 = 9 + 8 + 2 \times 3 \times 2\sqrt 2 = 17+12\sqrt 2$.

Rightly enough, $17^2 - 2 \times 12^2 = 1$.

EDIT : Only for completeness, I state what another answer above has also stated.

The point is, that once we write $a^2-nb^2 = (a+b\sqrt n)(a-b\sqrt n)$, we see that solutions of the equation $a^2-nb^2 = 1$ correspond to units of the ring $\mathbb Z[\sqrt n]$ (even when $n$ is negative!).

Thankfully, a part of algebraic number theory deals with this is detail. First, we note that there is a norm on $\mathbb Z[\sqrt n]$ given by $N(a+b\sqrt n) = a^2-nb^2$. This is a multiplicative function on this domain,hence one can check that the norm of an element is one if and only if it is a unit. Hence, the result boils down to which elements have norm one.

The Dirichlet Unit Theorem discusses the structure of the units of the ring of algebraic integers of a number field. More precisely, it states that the units are a product of a free abelian group of some order which can be calculated, and the roots of unity present in the number field. the rank of the free abelian group is the sum of the number of real conjugates(embeddings) and half the number of complex conjugates(embeddings), minus one.

For example, let us look at $\mathbb Z[\sqrt 2]$. $\sqrt 2$ has two conjugates, given by $\pm \sqrt 2$, both of which are real numbers. No complex conjugates exist. Hence, the rank of the finite abelian group is $2+0-1 = 1$. The roots of unity in $\mathbb Z[\sqrt 2]$ are just $\pm 1$. So, the group of units of $Z[\sqrt 2]$ is just ${\pm 1} \times \mathbb Z^1$. Now, the only question is about the generator of $\mathbb Z$. Usually, one can check small values for the answer : in our case, it is not difficult to check that $(1+\sqrt 2)$ has norm equal to $1$. So, all units in $\mathbb Z[\sqrt 2]$ are given by the coefficients of $\pm(1+\sqrt 2)^k$ for $k \in \mathbb Z$. Of course, note that we have some solutions above.

The theory of Pell's equations is therefore essentially a part of the study of quadratic number fields, as Wikipedia will also tell you.

  • 1
    $\begingroup$ Wouldn't (0,1) be the smallest solution? $\endgroup$ – WaveX Mar 25 '17 at 1:21
  • $\begingroup$ @WaveX You are right, but then the formula for further solutions would not work at all. You can check that for all $n$, $(1,0)$ is a solution to $x^2-ny^2 = 1$. So it qualifies (and I credit you for that), but only as a trivial solution. The attempt to search for non-trivial solutions (the smallest amongst these being the fundamental solution) is I think the key to recognizing whether or not the solution has infinitely many solutions or not. Having said that, thank you for pointing this out. $\endgroup$ – Teresa Lisbon Mar 25 '17 at 1:24
  • $\begingroup$ Your welcome. And I get what you're saying. That would make $(\pm2,\pm3)$ the first four non-trivial solutions to this specific case, correct? $\endgroup$ – WaveX Mar 25 '17 at 1:27
  • $\begingroup$ @WaveX Correct. $\endgroup$ – Teresa Lisbon Mar 25 '17 at 1:29

If $(a,b)$ is a solution to this equation, then $(2ab,4a^2+1)$ is as well:

$2(2ab)^2 + 1 = 8a^2b^2+1 = 8a^2(2a^2+1)+1 = 16a^4+8a^2+1 = (4a^2+1)^2$

And there is at least one solution : $(2,3)$. Hence, you will get infinitely many solutions: $(2*2*3=12,4*2^2+1=17)$, $(2*12*17=408,4*12^2+1=577)$, etc.


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