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.

 A: 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.
A: Suppose $a^2+b^2=c^2$ and $(a+1)^2+(b+1)^2=(c+1)^2$ both hold.
Simplifying the second equation and subtracting the first gives $$2a + 2b = 2c - 1$$ but this is impossible sso they can't both hold!
hint: think about even and odd numbers to complete the proof.
A: Since I assume that you're new to some Mathematical topics, I'd use basic language. 

First of all, the question itself teaches you that, if $(a,b,c)$ is a Pythagorean triplet, then it means that $a^2 + b^2 = c^2$. So if $\rm (foo_1, foo_2, foo_3)$ is a Pythagorean triplet, then $\rm (foo_1)^2 + (foo_2)^2 = (foo_3)^2$ (right?). So as you are given in the question, if $(a+1,b+1,c+1)$ is a triplet,then $(a + 1)^2 + (b + 1)^2 = (c + 1)^2$

You are left with two equations:$$\begin{align}  a^2 + b^2 &= c^2 \\ (a + 1)^2 + (b + 1)^2 &= (c + 1)^2\end{align}$$Subtract, and see the magic. Hint: knowledge of "even," and "odd" numbers will help. Especially that odd numbers are in the form $2m + 1 $ and even ones in $2m$... $m$  is an integer by the way.
A: Hint $\rm\: mod\ 2,\:$ where $\rm\:x^2 \equiv x,\:$ the equations imply $\rm\:\color{#C00}{c\!+\!1} \equiv a\!+\!1 + b\!+\!1\equiv a+b\equiv \color{#c00}c\ $ $\Rightarrow\Leftarrow$
Alternatively, $ $ if modular arithmetic is unfamiliar, then you can instead show that
$\rm\quad a^2\! +\! b^2\:$ and $\rm\:(a\!+\!1)^2\!+(b\!+\!1)^2\:$ have equal parity, but $\rm\:c^2$ and $\rm\,(c\!+\!1)^2$ have opposite parity
To do that, notice that squaring preserves parity, since $\rm\,odd^2\! = odd,\:$ $\rm\:even^2\! = even,\,$ hence the parity of $\rm\:(a\!+\!1)^2\! + (b\!+\!1)^2$ equals that of $\rm\:a\!+\!1 + b\!+\!1 = a+b+2\:$ equals that of $\rm\:a\!+\!b,\:$ since adding $2$ preserves parity. Similarly the parity of $\rm\:(c\!+\!1)^2\:$ equals that of $\rm\:c\!+\!1\:$ differs from $\rm\:c.\:$ Therefore we can't have $\rm\:(c\!+\!1)^2 = (a\!+\!1)^2\!+(b\!+\!1)^2\:$ since the LHS has parity same as $\rm\:c\!+\!1\:$ but, as shown, the RHS has parity same as $\rm\:a^2+b^2 = c^2\:$ same as $\rm\:c.$  
Remark $\ $ Notice how the above proof is shortened to a single line in my first proof, by arithmetizing it using  modular arithmetic. This  powerful and efficient  modular approach has widespread application in number theory (and algebra), so one should learn it as soon as possible.
A: 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. 
