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Find the number of pairs of positive integers (x, y) such

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Find the number of pairs of positive integers (x, y) such [#permalink] New post 11 Dec 2009, 21:23
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Find the number of pairs of positive integers (x, y) such that x^6 = y^2 + 127.

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4
[Reveal] Spoiler: OA

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Last edited by Bunuel on 03 Jul 2013, 00:13, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: x^6 = y^2 + 127 [#permalink] New post 11 Dec 2009, 23:06
xcusemeplz2009 wrote:
Find the number of pairs of positive integers (x, y) such that x^6 = y^2 + 127.

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4

pls share if u used a quick approach to solve this


First note that: x and y are +ve integers i.e. x and y cannot be -ves, zeros and fractions.

Given that:- x^6 = y^2 + 127
x can have a value of 1....................n, however x>2 because:

If x = 1, x^6 = 1 and y = sqrt(1-127). y is an irrational number. Not possible.
If x = 2, x^6 = 64 and y = sqrt(64-127). y is an irrational number. Not possible.
If x = 3, x^6 = 729 and y = sqrt(729-127) = sqrt(602). Now y is a fraction. Not possible.
If x = 4, x^6 = 64x64 and y = sqrt{(64x64) -127} = .......?

Now its not possible to do any calculation beyond this point in 2-5 minuets. Until and unless there is a quick approach, I would say this not a gmat-type question. I would love to see if anybody has a quick approach.

I thought its a good question however it turned to be a tough one.
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Re: x^6 = y^2 + 127 [#permalink] New post 12 Dec 2009, 00:48
Thats correct, even I used brute force. But can we find some better approach for similar question, where constant may be changed i.e. 64 or exponent may be changed. ?
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Re: x^6 = y^2 + 127 [#permalink] New post 12 Dec 2009, 06:16
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HI.. a quick appch i can think of..
the eq can be written as... x^6-y^2=127...x^6-y^2=(x^3-y)(x^3+y)....which means 127 is product of two int... but if u look at 127..it is prime number... so there is no pair of positive integers which satisfies the condition ...ANS-0
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Re: x^6 = y^2 + 127 [#permalink] New post 12 Dec 2009, 07:03
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xcusemeplz2009 wrote:
Find the number of pairs of positive integers (x, y) such that x^6 = y^2 + 127.

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4

pls share if u used a quick approach to solve this


chetan2u's approach above is excellent, though there's one problem with the analysis. We have

(x^3 + y)(x^3 - y) = 127

so x^3 + y and x^3 - y must be factors of 127. Since 127 is prime, the only possibility is that x^3 + y = 127 and x^3 - y = 1. Now, x cannot be greater than 5, since that would make x^3 larger than 127, and since x^3 must be greater than y, x could only be 4 or 5. Still testing these values, we do find that x = 4 and y = 63 gives a legitimate solution here, so there is one pair of values that works.
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Re: x^6 = y^2 + 127 [#permalink] New post 12 Dec 2009, 07:06
chetan2u wrote:
HI.. a quick appch i can think of..
the eq can be written as... x^6-y^2=127...x^6-y^2=(x^3-y)(x^3+y)....which means 127 is product of two int... but if u look at 127..it is prime number... so there is no pair of positive integers which satisfies the condition ...ANS-0


This is definitely a tough one... but I do not think the answer is 0.

chetan2u had a great approach, same as the one I used but since 127 is a prime number I went a step further:
(x^3-y)(x^3+y)=127
(x^3-y)(x^3+y)=(1)(127)

From here I plugged in different values of x and y to see if I can get (x^3-y) to equal 1 and (x^3+y) to equal 127. Because we know that (x^3-y) must equal 1, y must be x^3-1.

x=1, not possible
x=2, not possible
x=3, not possible

x=4, y=63
(x^3-y)(x^3+y)=127
(4^3-63)*(4^3+63)=127

x=5, not possible

So answer is B. 1

Can someone please verify the OA?
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Re: x^6 = y^2 + 127 [#permalink] New post 12 Dec 2009, 07:25
hi ianstewart...
thanks a lot... should not have skipped my mind but it seems in a hurry, just overlooked it...
lesson learnt-give a thought before proceeding to next step..
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Re: x^6 = y^2 + 127 [#permalink] New post 12 Dec 2009, 12:10
chetan2u wrote:
HI.. a quick appch i can think of..
the eq can be written as... x^6-y^2=127...x^6-y^2=(x^3-y)(x^3+y)....which means 127 is product of two int... but if u look at 127..it is prime number... so there is no pair of positive integers which satisfies the condition ...ANS-0


Thats a good point.

x^6-y^2 = 127
(x^3-y)(x^3+y) = 127 x 1

Since x and y both are +ves,
(x^3-y) = 1 and
(x^3+y) = 127

As I mentioned earlier, x >2. Now the possibilities for x are <5. So x could be 3 or 4 or 5.

After few trails, x = 4.

Agree with OA as B.
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Re: x^6 = y^2 + 127 [#permalink] New post 12 Dec 2009, 14:21
I believe, solving these kinds of questions is key to scoring well on the test. When we never know which question is experimental, a tough one as worse as this can always waste time and lower scores. I have had similar questions and I did not have sufficient practise facing them and lost valuable time. The key is to not panic and try to think at least one step beyond or differently for every 3 to 5 seconds when facing a tough problem.
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Re: x^6 = y^2 + 127 [#permalink] New post 21 Oct 2011, 02:34
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xcusemeplz2009 wrote:
Find the number of pairs of positive integers (x, y) such that x^6 = y^2 + 127.

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4

pls share if u used a quick approach to solve this


Here is what I would do:

First thing that comes to my mind is that 127 is prime. Well then, I can't do anything with it right now.
Then I see that there are some squares
(x^3)^2 - y^2 = 127
Let's sayx^3 = a

a^2 - y^2 = 127
(a+y)(a-y) = 127*1
We know that a and y are both positive integers. Therefore, their sum, a+y = 127 and their difference, a-y = 1.
It is obvious that a and y must be 64 and 63. (or you can solve the two equations simultaneously to get the values for a and y)
If a = 64 = x^3, x must be 4.
So there is only one pair of values (4, 63).

The question is straight forward because 127 is prime. You get only one pair of values. If instead, we have a composite number with many factors, we need to find the possible values of a and y and then see which values of a work for us.
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Re: Find the number of pairs of positive integers (x, y) such [#permalink] New post 28 Aug 2014, 23:05
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Re: Find the number of pairs of positive integers (x, y) such [#permalink] New post 05 Sep 2014, 07:38
Is it correct to assume that there can only be one set because the integers are positive and x's exponent is higher than y's? Without changing the equation you can almost see that there can only be one solution. But maybe I am thinking too simply...
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Re: Find the number of pairs of positive integers (x, y) such [#permalink] New post 08 Sep 2014, 21:44
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Is it correct to assume that there can only be one set because the integers are positive and x's exponent is higher than y's? Without changing the equation you can almost see that there can only be one solution. But maybe I am thinking too simply...


No. 127 is a prime number. Try putting in a composite number. Then see whether you get multiple values. Also, some other prime numbers such as 19 may give no solution.
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Re: Find the number of pairs of positive integers (x, y) such [#permalink] New post 10 Sep 2014, 01:01
IanStewart wrote:
xcusemeplz2009 wrote:
Find the number of pairs of positive integers (x, y) such that x^6 = y^2 + 127.

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4

pls share if u used a quick approach to solve this


chetan2u's approach above is excellent, though there's one problem with the analysis. We have

(x^3 + y)(x^3 - y) = 127

so x^3 + y and x^3 - y must be factors of 127. Since 127 is prime, the only possibility is that x^3 + y = 127 and x^3 - y = 1. Now, x cannot be greater than 5, since that would make x^3 larger than 127, and since x^3 must be greater than y, x could only be 4 or 5. Still testing these values, we do find that x = 4 and y = 63 gives a legitimate solution here, so there is one pair of values that works.


From above it would follow that if x & y are positive, x^3-y=1 & x^3+y=127. adding the two equations, 2x^3=128, x^3=64, x=4. since solvable with a unique solution, answer B
Re: Find the number of pairs of positive integers (x, y) such   [#permalink] 10 Sep 2014, 01:01
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