Consecutive Pythagorean Triples, Pythagoras Theorem Part2

Consecutive Pythagorean Triples methods from the history of math part:2

in the previous article: integer Pythagorean Triples, Methods From the History of Math Part 1,   we discussed a simple algorithm enables us to find integer triples; the current article introduces another method to find consecutive Pythagorean triples which means triples a,b,c
so that b (one right side) and c(the hypotenuse) are consecutive numbers.
The method based on the short multiplication formula:
(x+y)² = x²+2xy+y²
Suppose that y=1
We get the equation:
*** (x+1)²=x²+2x+1
In both sides we can notice squares of consecutive numbers which are: x and x+1
In order to receive a Pythagorean triple via this formula, the expression 2x+1 must be square of some alue.
In this case the parts of the equation: a2 + b2 = c2 that describes Pythagoras theorem are:
a² = 2x+1
b² = x²
c²= (x+1)²
lets suppose that 2x+1 is square of the value t.
then 2x+1=t² , therefore x=(t²-1)/2.
we can express the Pythagorean triple
a² = 2x+1 = t² ====== a=t
b² = x² ========== b=x ===== b=(t²-1)/2.
c²= (x+1)² ========== c=x+1 ===== c=(t²+1)/2 .
Note that if we wan to "build" integer consecutive Pythagorean triples, the value t must be odd number    ( try to figure out why??)
Now let us try to find Pythagorean triple using this template: lets chose for example 5 as the the value of t
Then:  a= t = 5
b= (25-1)/2 == b=12
c= (25+1)/2 == c=13
this lead to the consecutive triple (5,12,13), and using different odd values of t (t>1) leads us to different consecutive Pythagorean triples.