If $(a, b, c)$ is a Pythagorean triple, then $(a+1, b+1, c+1)$ is not a Pythagorean triple. Pythagoras stated that there exist positive natural numbers, $a$, $b$ and $c$ such that $a^2+b^2=c^2$. These three numbers, $a$, $b$ and $c$ are collectively known as a
Pythagorean triple. For example, $(8, 15, 17)$ is one of these triples as $8^2 + 15^2 = 64 + 225= 289 = 17^2$. Other examples of this triple are $(3, 4, 5)$ and $(5, 12, 13)$.


  Using Proof by Contradiction, show that: If $(a, b, c)$ is a Pythagorean triple, then $(a+1, b+1, c+1)$ is not a Pythagorean triple.
We must suppose that $(a+1,b+1,c+1)$ is in fact a Pythagorean triple, and that $(a,b,c)$ is, too. Then we have $$(a+1)^2+(b+1)^2=(c+1)^2\tag{1}$$ and $$a^2+b^2=c^2.\tag{2}$$

Expand $(1)$--using for example that $(a+1)^2=a^2+2a+1$--and then use $(2)$ to eliminate all the squared terms from the resulting equation. You should be able to conclude (after gathering the $a,b,c$ terms on one side and other terms on the other) that $1$ is an even number.
Suppose both statement are true. Then

$(a+1)^2+(b+1)^2=(c+1)^2$

and as well

$a^2+b^2=c^2$

From 1.

$a^2+2a+1+b^2+2b+1=c^2+2c+1$

$\implies 2a+2b=2c-1$

But this does not satisfy for Pythagorean triplet 3,4,5 or for any other like 5,12,13.

Hence our supposition is wrong.