Plethora of Pythagorean triples

if $a<b<c$ is a Pythagorean triple for odd $a$ , we know that $a^{2}=c^{2}-b^{2}=(b+c)(c-b)$

If we let $b,c$ be consecutive integers we get that $b+c=2b+1$ and $c-b=1$ .

Therefore we want to confirm that for every odd $a \geq 3$ , there exists a positive integer integer $b >a$ such that $a^{2}=2b+1$ .

Since $a^{2}$ is odd and $2b+1$ can be any odd number, we simply need to make sure that $b > a$ in all cases. $a^{2}$ grows faster that $2b+1$ for $3 \leq a<b$ , so if the smallest case works, they all work.

$3^{2}+4^{2}=5^{2}$ works, so all off $a \geq 3$ are the smallest member of a Pythagorean triple.

Therefore it is $\boxed{\text{True}}$

. Let $a$ be an odd positive integer greater than 1. Then $a^{2}+(\frac{a^{2}-1}{2})^{2}= (\frac{a^{2}+1}{2})^{2.}$

This stems from the "Pythagorean Triples Theorem."

If we use Euclid's formula for creating Pythagorean triples, but replace natural numbers $(m,n)$ where $m>n$ with $(m,m-1)$ we get that the first term usually denoted by $(m^2 - n^2)$ becomes the difference between two consecutive squares $(m^2 - (m-1)^2)$ which will simplify to the general term $2m-1$ for all natural numbers $m$ greater than or equal to 2, thus creating a smallest term that represents the set of all the odd natural numbers.

For every odd number $a$ there exist two perfect squares $x$ and $y$ such that $x+a=y$ , with $a<<x$ Let's consider the case where $y=x+1$ The above statement is true in particular for ann odd number $a$ that is also a perfect square. The initial statement then in demonstrated because gcm(x,x+1) is always 1.

Let us first look at c=(b+1)^2, this is b^2+2b+1, 2b+1 describes any odd number for integer b, so this odd number could be the square of any odd number. Let us say this odd number, our odd number, is a=(2b+1)^0.5, we solve for b=((a^2)-1)/2, if we add 1 to get the sum in the Pythagorean triple we get c=((a^2)+1)/2. Now note that c=b+1 so these numbers have a gcd of 1, also do note that a^2 is a multiple of a, so (a^2)+1 and (a^2)-1, both of which have a difference of one, also have a gcd of 1 with a, and since the prime factorisation of the numbers is fundamentally different (no shared primes), and dividing by two would only take away primes (2) without adding primes, a shares no common factors with b or c above 1. We hence have an example where the gcd is one for every odd value above 1.

we can take the cases of --- 1st-3,4&5 2nd-5,12&13 and so on......