Pythagoras' Theorem For any right-angled triangle: $a^2 + b^2 = c^2$ where $a$, $b$ are t

Question

Pythagoras' Theorem For any right-angled triangle: $a^2 + b^2 = c^2$ where $a$, $b$ are the lengths of the two shorter sides, and $c$ is the hypotenuse. Pythagorean Triples • A set of three positive integers $(a,b,c)$ that satisfy $a^2 + b^2 = c^2$ are called Pythagorean Triples. • Examples: $(3,4,5)$, $(5,12,13)$, $(6,8,10)$, $(7,24,25)$, $(8,15,17)$, etc. Generate Pythagorean Triples using an odd number (except $1$): Given any odd number, $2n+1$, where $n=1,2,3,...$, the square of $2n+1$ can be expressed as the sum of two consecutive numbers, such that $(2n+1)^2 = k + (k+1)$. Using an algebraic method, or otherwise, show that $(2n+1, k, k+1)$ is a set of Pythagorean Triples. Generate Pythagorean Triples using an even number (except $2$): Given any even number, $2n$, where $n=2,3,...$, then the square of a half of $2n$ is $n^2$. Consider the other two numbers that are one less than and one more than $n^2$, that is, $n^2 - 1$ and $n^2 + 1$. Using an algebraic method, or otherwise, show that $(2n, n^2 - 1, n^2 + 1)$ is a set of Pythagorean Triples.

Accepted Answer

(a) $k=2n(n+1)$ and $(2n+1)^2+k^2=(k+1)^2$. (b) $(2n)^2+(n^2-1)^2=(n^2+1)^2$.

Pythagorean Triples

Pythagorean Triples

Solution:

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