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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.

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Welcome to math.SE. Since you're new here, let me give you a few pieces of advice. For one, you are asking us for help, not giving us an assignment. It would be nice if you formulated your question to reflect that fact. Second, tell us where you're stuck. That way you help us help you understand better. – Arthur Nov 17 '12 at 16:11
yes..sorry,my bad – niki Nov 17 '12 at 16:43
excuse me,can you give me sample answer for this question?urgent – niki Nov 17 '12 at 16:51
You've got two answers already, niki. – Cameron Buie Nov 17 '12 at 16:52
More generally, if (a,b,c) is a Pythagorean triple and k is a positive integer, then (a+k,b+k,c+k) is not a Pythagorean triple. – Did Nov 17 '12 at 17:28

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.

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Magic point at the last! 1 is not an even number. Easy one. – Babak S. Nov 17 '12 at 16:19

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.

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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.

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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.

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Suppose both statement are true. Then


and as well


From 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.

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I don't understand your concluding statement regarding your arrived contradiction. – Jonathan Hebert Jun 1 '15 at 14:13
I may have put in weak words to support that. The equation $2a+2b=2c-1$ will not hold true for any 3 given values, that are already a pythagorean triplet, that flaws our supposition about equation 1. – MonK Jun 1 '15 at 14:18
Why won't it hold true for a Pythagorean triple (a,b,c)? – Jonathan Hebert Jun 1 '15 at 17:01

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