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# If b is an integer, is sqr root (a^2 + b^2) an integer? 1)

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If b is an integer, is sqr root (a^2 + b^2) an integer? 1) [#permalink]  15 Apr 2006, 19:18
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If b is an integer, is sqr root (a^2 + b^2) an integer?

1) a^2 + b^2 is an integer.
2) a^2 - 3b^2 = 0
Manager
Joined: 08 Feb 2006
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B for me

1)square root of integer, still don't know if it is an integer or not

2)a^2 -3b^2
= a^2=3b^2

squareroot 3b^2+b^2
= squareroot 4b^2
=2b
Sufficient since b is an integer
VP
Joined: 29 Dec 2005
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jlui4477 wrote:
B for me
1)square root of integer, still don't know if it is an integer or not
2) a^2 -3b^2 = a^2=3b^2

= squareroot 3b^2+b^2
= squareroot 4b^2
= 2b
Sufficient since b is an integer

= squareroot 4b^2 = + or - 2b (but still is an integer)

B.
Manager
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Thanks for the correction
VP
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Another one for B!

A doesnt work!

B:

a^2 + b^2 = a^2 - 3B^2 + 4b^2 = 0 + 4B^2

So sqrt(a^2 + b^2) = 2b! Which is an integer!
Manager
Joined: 20 Nov 2004
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Professor wrote:
= squareroot 4b^2 = + or - 2b (but still is an integer)
B.

Professor,

it's true that (+2b)^2 and (-2b)^2 lead to 4b^2, but
the square root of 4b^2 is by definition only +2b.
VP
Joined: 29 Dec 2005
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huuuuuuuuhhhhhhh............

i take back my statement... you are correct........

ccax wrote:
Professor wrote:
= squareroot 4b^2 = + or - 2b (but still is an integer)
B.

Professor,

it's true that (+2b)^2 and (-2b)^2 lead to 4b^2, but
the square root of 4b^2 is by definition only +2b.
Manager
Joined: 27 Mar 2006
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Re: DS [#permalink]  18 Apr 2006, 09:29
john2005 wrote:
If b is an integer, is sqr root (a^2 + b^2) an integer?

1) a^2 + b^2 is an integer.
2) a^2 - 3b^2 = 0

B it is
Re: DS   [#permalink] 18 Apr 2006, 09:29
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