Generating Sequence of Right angled triangle

Question

A "nearly isoceles" right-angled triangle with integer side lengths is defined as one in which the two sides adjacent to the right angle differ in length by just 1 unit. A triangle with side lengths 20,21 and 29 is an example. Find with proof , a method for generating a sequence of such triangles. I do not have my work because I cannot understand that how can I approach this problem or what is the strategy to solve it. Any help would be appreciated.

Answer 1

Remember that a Pythagorean triangle with relatively prime legs has sides $2mn, m^2-n^2, m^2+n^2 $ for some integers $m$ and $n$.

If the legs are $a$ and $a+1$, since they are relatively prime, the legs must satisfy either $a=2mn, a+1=m^2-n^2$ or $a=m^2-n^2, a+1=2mn$.

Find the solutions to these equations and you will have your triangles.

Answer 2

$\textbf{Matching the difference} \quad X=|B-A|$

There are $\space3\space$ approaches we can take for this:

$X$ can be any prime number $(p)$ where $p\equiv\pm1\mod 8$ with (p) raised to any non-negative power. Under $100$, $X\in \{1,7, 17, 23, 31, 41, 47, 49, 71, 73, 79, 89, 97\}$.

1. Beginning with $X=1$, a solution was provided by Wacław Sierpiński $\quad$[Pythagorean triangles, THE SCRPTA MATHEMATICA STUDIES Number NINE, GRADUATE SCHOOL OF SCIENCE YESHIVA UNIVERSITY, NEW YORK, 1962, pp. 17-22 ]$\quad$ with equations that resolve to the following for generating these triples $(T_n)$ in sequence with a starting "seed" of $T_0=(0,0,1)$. \begin{equation} A_{n+1}=3A_n+2C_n+1\\ B_{n+1}=3A_n+2C_n+2 \\ C_{n+1}=4A_n+3C_n+2\\ \ T_1=(3,4,5)\quad T_2=(20,21,29)\\ T_3=(119,120,168)\quad T_4=(696,697,985) \end{equation}

We can use Euclid's formula shown here as $$ A=m^2=k^2\quad B=2mk\quad C=m^2+k^2 $$

2. and we can find the $\space (m,k)\space$ numbers to feed the formula with the following equation where $\space k\space$ begins at $\space1\space$ and each $\space m\space$ found becomes $\space k\space$ for the next iteration. $$m=k+\sqrt{2k^2+(-1)^k}$$ \begin{align*} k=1\space& m=1+\sqrt{2(1)^2+(-1)^1}=2 & F(2,1)=(3,4,5)\\ k=2 \space& m=2+\sqrt{2(2)^2+(-1)^2}=5 & F(5,2)=(21,20,29)\\ k=5 \space& m=5+\sqrt{2(5)^2+(-1)^5}=12 & F(12,5)=(119,120,169)\\ k=12 \space& m=12+\sqrt{2(12)^2+(-1)^{12}}=29 & F(29,12)=(697,696,985) \end{align*}

The series $\{0, 1, 2, 5, 12, 29, 70, 169, ...\}$ are Pell numbers

3. and the $(m,k)$-values needed of them may be generated directly using the following: $$ m_n= \frac{(1 + \sqrt{2})^{n+1} - (1 - \sqrt{2})^{n+1}}{2\sqrt{2}}\\ k_n= \frac{(1 + \sqrt{2})^n - (1 - \sqrt{2})^n}{2\sqrt{2}}$$ For example \begin{align*} m_1= \frac{(1 + \sqrt{2})^{2} - (1 - \sqrt{2})^{2}}{2\sqrt{2}}=2 & \quad k_1= \frac{(1 + \sqrt{2})^1 - (1 - \sqrt{2})^1}{2\sqrt{2}}=1 \\& F(2,1)=(3,4,5)\\ m_2= \frac{(1 + \sqrt{2})^{3} - (1 - \sqrt{2})^{3}}{2\sqrt{2}}=5& \quad k_2= \frac{(1 + \sqrt{2})^2 - (1 - \sqrt{2})^2}{2\sqrt{2}}=2 \\& F(5,2)=(21,20,29)\\ m_3= \frac{(1 + \sqrt{2})^{4} - (1 - \sqrt{2})^{4}}{2\sqrt{2}}=12& \quad k_3= \frac{(1 + \sqrt{2})^3 - (1 - \sqrt{2})^3}{2\sqrt{2}}=5 \\& F(12,5)=(119,120,169)\\ m_4= \frac{(1 + \sqrt{2})^{5} - (1 - \sqrt{2})^{5}}{2\sqrt{2}}=29& \quad k_4= \frac{(1 + \sqrt{2})^4 - (1 - \sqrt{2})^4}{2\sqrt{2}}=12 \\& F(29,12)=(697,696,985) \end{align*}