Congruent Numbers

Over the years I've sometimes made note of when a number associated with my diurnal age is congruent. Often I've just ignored the fact. The definition of such number is:

A number is called congruent it is the area of a right triangle with rational sides.

The first congruent number is 5 because it is the area of a right triangle with sides of 20/3, 3/2, and 41/6. The next number, 6, is also congruent being the area of the famous 3, 4, 5 right triangle. See Figure 1.

Figure 1

7 is also a congruent number being the area of a right triangle with sides such that (see Figure 2):$$ \begin{align} \Big ( \frac{35}{12} \Big )^2+ \Big ( \frac{24}{5} \Big )^2 &= \Big ( \frac{337}{60} \Big )^2 \\ \frac{1}{2} \cdot \frac{35}{12} \cdot \frac{24}{5} &= 7 \end{align}$$

Figure 2

With 7, we begin to see the problem of determining whether a number is congruent or not. How do we find those fractions that confirm that 7 is congruent. It's not easy. To quote from Wikipedia:

The question of determining whether a given rational number is a congruent number is called the congruent number problem. This problem has not (as of 2019) been brought to a successful resolution. Tunnell's theorem provides an easily testable criterion for determining whether a number is congruent; but his result relies on the Birch and Swinnerton-Dyer conjecture, which is still unproven.

Without getting too deeply into this topic, it suffices to say that OEIS A006991 provides a list of so-called primitive congruent numbers up to 10,000.  These are congruent numbers that are square-free. If any of these numbers are multiplied by a positive square number, a new congruent number is created.

Let's take the number associated with my diurnal age yesterday as an example. 27260 factorises as follows:$$ \begin{align} 27260 &= 2^2 \times 5 \times 29 \times 47\\ &= 2^2 \times 6815 \end{align}$$Now 6815 is a member of OEIS A006991 and therefore 27260 is thus a congruent number because a primitive congruent number, 6815, has been multiplied by a positive square number (4).

It can be noted that 6814 is also a member of OEIS A006991 and thus 4 x 6814 = 27256 is also a congruent number. It's not uncommon for primitive congruent numbers to occur in pairs or even triples.  For example: (5, 6, 7), (13, 14, 15), (21, 22, 23), (29, 30, 31), 34 is a singleton, (37, 38, 39) etc.

Rational Solutions to a^a = b^b

Here's a problem that I encountered on this site:

Find all pairs of rational numbers \((a,b)\) such that \(a^a=b^b\) for \(0 < a < b\).

One solution that was proposed runs as follows:

Write \(b=ax\) for some \(x>1\). Then \(x=b/a\) so \(x\) is rational.

Then taking logs, we get $$ \begin{align} a\log a&=b\log b\\&=ax\log(ax) \\ \log a&= x \log(ax) \\&=x \log(x)+x \log(a)      \\&=\frac{x\log x}{1-x} \\  a&=x^{\frac{x}{1-x}}\\b&=x^{\frac{1}{1-x}} \end{align} \\ \text{since } x\neq1 \text{ and } a\neq0$$Since \(x\) is rational, it suffices to find the values of \(x>1\) for which \(x^{\frac{1}{1-x}}\) is rational.

We claim that the only such values of \(x\) are \(x=\dfrac{n+1}{n}\) where \(n\) is an integer.

We may write \(x=1+\dfrac1n\) where \(n\in\mathbb{Q}\), \(n>0\).

It is easy to see if \(n\) is a positive integer that \(x^{\frac{1}{1-x}}\) is rational.

If \(n\) is not an integer, write \(n=p/q\) where \(p,q\in\mathbb{Z}^+\) are coprime with \(q>1\).

We have \(x^{\frac{1}{1-x}}=\left(\dfrac{n}{n+1}\right)^n=\left(\dfrac{p}{p+q}\right)^{p/q}\).

Hence \(p\) and \(p+q\) must be \(q \,\)th powers.

No two \(q \,\)th powers can differ by \(q\) since for positive integers \(u,v,q\), we have by the Binomial Theorem, $$(u+v)^q-u^q=qu^{q-1}v+\ldots+v^q>q\,.$$Therefore, if \(n\) is not an integer, \(x^{\frac{1}{1-x}}\) is not rational, so the only rational solutions are given by $$a=\left(\dfrac{n}{n+1}\right)^{n+1} \text{ and } \, b=\left(\dfrac{n}{n+1}\right)^n \text{ with } n\in\mathbb{Z}^+$$The initial values of \(a\) and \(b\) are shown in Figure 1.

Figure 1: permalink

Continued Fractions of Square Roots

It's all too easy, using online resources like WolframAlpha or SageMathCell, to extract a variety of information about continued fractions. For example, today I turned \(26195\) days old and this number is a member of OEIS A042431:

A042431

Denominators of continued fraction convergents to \( \sqrt{743} \).

To confirm that \(26195\) is indeed a member of this sequence, I would normally use SageMathCell but others would turn to WolframAlpha and, using the latter, it can be seen in Figure 1 that \(26195\) is indeed a denominator in one of the convergent fractions, specifically:$$ \displaystyle \frac{714024}{26195}$$

Figure 1

It's easy to take for granted that the continued fraction of an irrational number that is a square root (like \( \sqrt{743} \)) is always periodic. A good explanation of why this is so can be found on this site. In the following, I'm reproducing the site's conversion of \( \sqrt{5} \) to a continued fraction (and practising my LaTeX at the same time).

$$\begin{aligned}
\sqrt{5}&=2+x\\
5&=(2+x)^2 \\
&=4+4x+x^2\\
&=4+x\,(4+x)\\
5-4&=x\,(4+x)\\
1&=x\,(4+x)\\
x&=\frac{1}{4+x}\\
\text{ Thus } \sqrt{5}&=2+\frac{1}{4+x}\\
&=2+\frac{1}{4+\displaystyle \frac{1}{4+x}}\\
&=2+\frac{1}{4+\displaystyle \frac{1}{4+\displaystyle \frac{1}{4+x}}} \text{ etc.}\\
\end{aligned}$$Clearly, we can write \( \sqrt{5}=[2;\overline 4]\) where the overline represents repetition. In the case of \( \sqrt{743}\), we have \([27;\overline{3,1,7,27,7, 1, 3, 54}]\). The site referred to earlier goes on to develop a general algorithm for determining the periodic continued fraction of any irrational square root. The table at the end of this post lists the continued fractions of the square roots of the first 99 natural numbers (even those that aren't irrational). 

√n [ a; Period ] √1 [ 1; ] √2 [ 1; 2 ] √3 [ 1; 1, 2 ] √4 [ 2; ] √5 [ 2; 4 ] √6 [ 2; 2, 4 ] √7 [ 2; 1, 1, 1, 4 ] √8 [ 2; 1, 4 ] √9 [ 3; ] √10 [ 3; 6 ] √11 [ 3; 3, 6 ] √12 [ 3; 2, 6 ] √13 [ 3; 1, 1, 1, 1, 6 ] √14 [ 3; 1, 2, 1, 6 ] √15 [ 3; 1, 6 ] √16 [ 4; ] √17 [ 4; 8 ] √18 [ 4; 4, 8 ] √19 [ 4; 2, 1, 3, 1, 2, 8 ] √20 [ 4; 2, 8 ] √21 [ 4; 1, 1, 2, 1, 1, 8 ] √22 [ 4; 1, 2, 4, 2, 1, 8 ] √23 [ 4; 1, 3, 1, 8 ] √24 [ 4; 1, 8 ] √25 [ 5; ] √26 [ 5; 10 ] √27 [ 5; 5, 10 ] √28 [ 5; 3, 2, 3, 10 ] √29 [ 5; 2, 1, 1, 2, 10 ] √30 [ 5; 2, 10 ] √31 [ 5; 1, 1, 3, 5, 3, 1, 1, 10 ] √32 [ 5; 1, 1, 1, 10 ] √33 [ 5; 1, 2, 1, 10 ] √34 [ 5; 1, 4, 1, 10 ] √35 [ 5; 1, 10 ] √36 [ 6; ] √37 [ 6; 12 ] √38 [ 6; 6, 12 ] √39 [ 6; 4, 12 ] √40 [ 6; 3, 12 ] √41 [ 6; 2, 2, 12 ] √42 [ 6; 2, 12 ] √43 [ 6; 1, 1, 3, 1, 5, 1, 3, 1, 1, 12 ] √44 [ 6; 1, 1, 1, 2, 1, 1, 1, 12 ] √45 [ 6; 1, 2, 2, 2, 1, 12 ] √46 [ 6; 1, 3, 1, 1, 2, 6, 2, 1, 1, 3, 1, 12 ] √47 [ 6; 1, 5, 1, 12 ] √48 [ 6; 1, 12 ] √49 [ 7; ] √50 [ 7; 14 ]

√n [ a; Period ] √51 [ 7; 7, 14 ] √52 [ 7; 4, 1, 2, 1, 4, 14 ] √53 [ 7; 3, 1, 1, 3, 14 ] √54 [ 7; 2, 1, 6, 1, 2, 14 ] √55 [ 7; 2, 2, 2, 14 ] √56 [ 7; 2, 14 ] √57 [ 7; 1, 1, 4, 1, 1, 14 ] √58 [ 7; 1, 1, 1, 1, 1, 1, 14 ] √59 [ 7; 1, 2, 7, 2, 1, 14 ] √60 [ 7; 1, 2, 1, 14 ] √61 [ 7; 1, 4, 3, 1, 2, 2, 1, 3, 4, 1, 14 ] √62 [ 7; 1, 6, 1, 14 ] √63 [ 7; 1, 14 ] √64 [ 8; ] √65 [ 8; 16 ] √66 [ 8; 8, 16 ] √67 [ 8; 5, 2, 1, 1, 7, 1, 1, 2, 5, 16 ] √68 [ 8; 4, 16 ] √69 [ 8; 3, 3, 1, 4, 1, 3, 3, 16 ] √70 [ 8; 2, 1, 2, 1, 2, 16 ] √71 [ 8; 2, 2, 1, 7, 1, 2, 2, 16 ] √72 [ 8; 2, 16 ] √73 [ 8; 1, 1, 5, 5, 1, 1, 16 ] √74 [ 8; 1, 1, 1, 1, 16 ] √75 [ 8; 1, 1, 1, 16 ] √76 [ 8; 1, 2, 1, 1, 5, 4, 5, 1, 1, 2, 1, 16 ] √77 [ 8; 1, 3, 2, 3, 1, 16 ] √78 [ 8; 1, 4, 1, 16 ] √79 [ 8; 1, 7, 1, 16 ] √80 [ 8; 1, 16 ] √81 [ 9; ] √82 [ 9; 18 ] √83 [ 9; 9, 18 ] √84 [ 9; 6, 18 ] √85 [ 9; 4, 1, 1, 4, 18 ] √86 [ 9; 3, 1, 1, 1, 8, 1, 1, 1, 3, 18 ] √87 [ 9; 3, 18 ] √88 [ 9; 2, 1, 1, 1, 2, 18 ] √89 [ 9; 2, 3, 3, 2, 18 ] √90 [ 9; 2, 18 ] √91 [ 9; 1, 1, 5, 1, 5, 1, 1, 18 ] √92 [ 9; 1, 1, 2, 4, 2, 1, 1, 18 ] √93 [ 9; 1, 1, 1, 4, 6, 4, 1, 1, 1, 18 ] √94 [ 9; 1, 2, 3, 1, 1, 5, 1, 8, 1, 5, 1, 1, 3, 2, 1, 18 ] √95 [ 9; 1, 2, 1, 18 ] √96 [ 9; 1, 3, 1, 18 ] √97 [ 9; 1, 5, 1, 1, 1, 1, 1, 1, 5, 1, 18 ] √98 [ 9; 1, 8, 1, 18 ] √99 [ 9; 1, 18 ]