Babenzgs

2022-01-29

Integer between the 's p and $p+2$ is a perfect square. Explain why there exists an integer n such that

$4{n}^{2}-1=p$

a) Is the integer between p and$p+2$ odd or even?

b) Assume additionally that the integer between the 's p and$p+2$ is a perfect square.

c) By considering$(2n-1)(2n+1)$ , find the only possible value of p.

a) Is the integer between p and

b) Assume additionally that the integer between the 's p and

c) By considering

ocretz56

Beginner2022-01-30Added 16 answers

Step 1

a) If$p=2$ then

$p+2=4$
is not ,

a) If

Eliza Norris

Beginner2022-01-31Added 15 answers

Step 1

Given:$A}^{2}+{B}^{2}={C}^{2$ the most common means of generating Pythagorean triples is Euclids

Given: