Few and Far Between?

 

We designed an Excel programme in order to find the examples that the smaller triples differ by one so that we could, hopefully, find the pattern of them and find out a general solution of them. Considering the 65534 numbers, which is the maximum number of rows that Excel provides, we have found that there are 7 sets of solutions.

 

Using Excel, I have found that there are six such triples with the smallest number below 100,000. I did this by calculating $\sqrt{n^2 + (n+1)^2}$ and looking for integers (using integer n's).

 

The triples are:

3,4,5;

20,21,29;

119,120,169;

696,697,985;

4059,4060,5741;

23660,23661,33461

 

Mahdokht from Farzanegan of Kermanshah used the formula which generates all Pythagorean triples which don't share a common factor: If $a^2+b^2=c^2$ then there are natural numbers $m> n\geq 1$ for which $$a=mn, b=\frac{m^2-n^2}{2}, c=\frac{m^2+n^2}{2}$$ Mahdokht then investigated solving $b=a+1$ and found the equation
$$m^2-2mn -n^2-2=0$$ The solution is $$m = n \pm \sqrt{2n^2+2}$$ and, as Mahdokht points out, this mean that $2n^2+2$ must be a square number. Using this as a guide, Mahdokht found the triples 20, 21, 29 and 119, 120, 169.

 

Eleanor from Dr. Challoner's High School made a wonderful observation in her solution:

By beginning with the obvious statement $n^2+(n+1)^2=(n+a)^2$, with $n$ and $a$ both in the natural numbers, it is possible to immediately reduce the statement to an equation for $n$ in $a$. $$n^2+(n+1)^2=(n+a)^2 \Leftrightarrow n=a-1\pm \sqrt{2a(a-1)}$$

 

It is immediately obvious that any $a$ which is a natural number > 1 and for which $\sqrt{2a(a-1)}$ is also a natural number  generates a value $n$ which has the desired property. By rewriting the square root as $2a(a-1)=b^2$, where $b$ is an integer, a second quadratic equation can be realised: $$2a^2-2a-b^2=0 \Leftrightarrow a=\frac{1 \pm \sqrt{2b^2+1}}{2}$$ By letting $(b+c)^2=2b^2+1$, with $c$ in the integers, as above, a third quadratic equation can be formed: $$(b+c)^2=2b^2+1 \Leftrightarrow b=c \pm \sqrt{2c^2-1}$$ Repeating this process again, defining $(c+d)^2=2c^2-1$ results in the quadratic equation: $$c=d \pm \sqrt{2d^2+1}$$ Repeating this any more times causes the sign of the $1$ in the square root to alternate, but otherwise results in the same equation, with different (and quickly running out) variable names. This can be interpreted as a sort of recurrence relation. It is also not a terrible idea to make clear that there is presently no guarantee that an integer term in this relation will generate another integer term. Due to the context of the values (ie. their intended use), the negative values should be disregarded completely, as they simply generate duplicate triples. Similarly, the $(0,1,1)$ triple has been disregarded, as it is a trivial triple, and hence of no real significance. Hence, let $z_{x+1}=z_x+\sqrt{2z_x^2+(-1)^x}$ and $z_1=2$, $2$ being the value which generates the $(3,4,5)$ triple. It is clearly now the case that $z_{2x+1}$ will generate a Pythagorean triple (and, more specifically, a primitive triple with a difference of 1 between the smaller legs), provided that $z_{2x+1}$ and $\sqrt{2z_{2x+1}^2+1}$ are both integers, which, due to the definition of $z_x$ , indicates that it is necessary and sufficient that $z_{2x+1}$ and $z_{2x+2}$ are integers.

 

After a small amount of algebraic manipulation, it may be seen that

$$2z_{x+1}^2+(-1)^x=(2z_x+\sqrt{2z_x^2+(-1)^{x+1}})^2,$$ and this can be used as the basis for mathematical induction on $\sqrt{2z_{x+1}^2+(-1)^x}$. Use $z_1=2, 2z_1^2+1=9$ as the base case for the aforementioned induction, and observe that $\sqrt{2z_{x+1}^2+(-1)^x}=2z_x+\sqrt{2z_x^2+(-1)^{x+1}}$.

 

Hence, if $\sqrt{2z_x+(-1)^{x+1}}$ is an integer, then $z_x$ is also an integer and so is $\sqrt{2z_{x+1}^2+(-1)^n}$. Because the base case is true, and every preceding case's truth confirms its successor's truth, it is instantly seen that every term in the sequence is an integer (and, hence, every odd term in the sequence is a candidate value for $b$ and can be used to generate a triple). Finally, by simplifying the quadratic equations out from the beginning of the process, it can be seen that $n_x=z_{2x-1}+\frac{\sqrt{2z_{2x-1}^2+1}-1}{2}$, and hence that $n_x^2+(n_x+1)^2=(z_{2x-1}+\sqrt{2z_{2x-1}^2+1})^2$.

 

As it has already been proven that $\sqrt{2z_{2x-1}^2+1}$ is an integer for all $x$, this is an additional confirmation that the solutions generated are valid. This also describes the relationship between the longest leg of the triple and the even-indexed $z$ values: $$n_x^2+(n_x+1)^2=z_{2x}$$ From this, it is possible to create each triple from the previous, starting from the smallest and working up infinitely. This, as well as other results, prove that there are infinitely many triples where the shorter legs differ by 1. However, as the sequence was not shown as the only one that generates such triples, it is possible that not all triples with this property will be generated.