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If x and y are integers, is x^2-y^2 odd?

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If x and y are integers, is x^2-y^2 odd?  [#permalink]

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New post 06 Feb 2019, 15:40
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GMATH practice exercise (Quant Class 16)

If \(x\) and \(y\) are integers, is \(\,{x^2} - {y^2}\,\) odd?

\(\left( 1 \right)\,\,\,x + y\,\,\) is odd
\(\left( 2 \right)\,\,\,x - y\,\,\) is odd

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Re: If x and y are integers, is x^2-y^2 odd?  [#permalink]

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New post 06 Feb 2019, 23:46
fskilnik wrote:
GMATH practice exercise (Quant Class 16)

If \(x\) and \(y\) are integers, is \(\,{x^2} - {y^2}\,\) odd?

\(\left( 1 \right)\,\,\,x + y\,\,\) is odd
\(\left( 2 \right)\,\,\,x - y\,\,\) is odd



Key word : \(x\) and \(y\) are integers, can be -ive or +ive

x^2 - y^2 = x-y x+y

So now the product of 2 expressions to be odd, both expressions need to be odd, said that

from 1) x + y is odd
lets put back in the question, this means that x-y will be odd as well, Even + odd = Odd, Even - Odd = Odd

from 2) x - y is odd
lets put back in the question, this means that x+y will be odd as well, Even - odd = Odd, odd + even= Odd

D
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Re: If x and y are integers, is x^2-y^2 odd?  [#permalink]

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New post 07 Feb 2019, 06:18
fskilnik wrote:
GMATH practice exercise (Quant Class 16)

If \(x\) and \(y\) are integers, is \(\,{x^2} - {y^2}\,\) odd?

\(\left( 1 \right)\,\,\,x + y\,\,\) is odd
\(\left( 2 \right)\,\,\,x - y\,\,\) is odd

\(x,y\,\,{\rm{ints}}\,\,\,\,\,\left( * \right)\)

\({x^2} - {y^2}\,\,\mathop = \limits^? \,\,{\rm{odd}}\)


First Approach: ("the smart way")

\(\left( 1 \right)\,\,x + y\,\,{\rm{odd}}\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,x,y\,\,:\,\,\,{\rm{one}}\,\,{\rm{odd}}\,{\rm{,}}\,\,{\rm{another}}\,\,{\rm{even}}\,\,\,\,\, \Rightarrow \,\,\,\,x - y\,\,{\rm{odd}}\,\,\,\,\, \Rightarrow \,\,\,\,{\rm{SUFF}}.\,\,\,\,\left( {**} \right)\)

\(\left( 2 \right)\,\,x - y\,\,{\rm{odd}}\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,x,y\,\,:\,\,\,{\rm{one}}\,\,{\rm{odd}}\,{\rm{,}}\,\,{\rm{another}}\,\,{\rm{even}}\,\,\,\,\, \Rightarrow \,\,\,\,x + y\,\,{\rm{odd}}\,\,\,\,\, \Rightarrow \,\,\,\,{\rm{SUFF}}.\,\,\,\,\left( {**} \right)\)


\(\left( {**} \right){x^2} - {y^2} = \left( {x + y} \right)\left( {x - y} \right) = {\rm{odd}} \cdot {\rm{odd}} = {\rm{odd}}\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{YES}}} \right\rangle\)


Second Approach: ("the elegant way")

\(\left( 1 \right)\,\,\,x + y\,\, = {\rm{odd}}\,\,\,\,\, \Rightarrow \,\,\,\,x - y = \underbrace {x + y}_{{\rm{odd}}} - \underbrace {\,2y\,}_{\left( * \right)\,\,{\rm{even}}} = {\rm{odd}}\,\,\,\,\, \Rightarrow \,\,\,\,{\rm{SUFF}}.\,\,\,\,\left( {***} \right)\)

\(\left( 2 \right)\,\,\,x - y\,\, = {\rm{odd}}\,\,\,\,\, \Rightarrow \,\,\,\,x + y = \underbrace {x - y}_{{\rm{odd}}} + \underbrace {\,2y\,}_{\left( * \right)\,\,{\rm{even}}} = {\rm{odd}}\,\,\,\,\, \Rightarrow \,\,\,\,{\rm{SUFF}}.\,\,\,\,\left( {***} \right)\)


\(\left( {***} \right){x^2} - {y^2} = \left( {x + y} \right)\left( {x - y} \right) = {\rm{odd}} \cdot {\rm{odd}} = {\rm{odd}}\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{YES}}} \right\rangle\)



The correct answer is therefore (D).


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net
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Re: If x and y are integers, is x^2-y^2 odd?   [#permalink] 07 Feb 2019, 06:18
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