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16 Sep 2014, 01:26
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If \(x\) and \(y\) are integers, is \(y\) an even integer?
(1) \(4y^2+3x^2=x^4+y^4\)
(2) \(y=4-x^2\)
(1) \(4y^2+3x^2=x^4+y^4\)
(2) \(y=4-x^2\)
16 Sep 2014, 01:26
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Official Solution:
If \(x\) and \(y\) are integers, is \(y\) an even integer?
(1) \(4y^2+3x^2=x^4+y^4\)
Rearrange: \(3x^2-x^4=y^4-4y^2\);
Factor out: \(x^2(3-x^2)=y^2(y^2-4)\).
Note that the left-hand side (LHS) is always even, irrespective of \(x\): if \(x\) is odd, then \(3-x^2=odd-odd=even\), and if \(x\) is even, then the product is naturally even. Therefore, \(y^2(y^2-4)\) on the right-hand side (RHS) must also be even. For this to hold, \(y\) must be even. If \(y\) were odd, then \(y^2(y^2-4)=odd*(odd-even)=odd*odd=odd\), contradicting the fact that RHS is even. Sufficient.
(2) \(y=4-x^2\)
If \(x\) is odd, then \(y=even-odd=odd\), but if \(x\) is even, then \(y=even-even=even\). Not sufficient.
Answer: A
If \(x\) and \(y\) are integers, is \(y\) an even integer?
(1) \(4y^2+3x^2=x^4+y^4\)
Rearrange: \(3x^2-x^4=y^4-4y^2\);
Factor out: \(x^2(3-x^2)=y^2(y^2-4)\).
Note that the left-hand side (LHS) is always even, irrespective of \(x\): if \(x\) is odd, then \(3-x^2=odd-odd=even\), and if \(x\) is even, then the product is naturally even. Therefore, \(y^2(y^2-4)\) on the right-hand side (RHS) must also be even. For this to hold, \(y\) must be even. If \(y\) were odd, then \(y^2(y^2-4)=odd*(odd-even)=odd*odd=odd\), contradicting the fact that RHS is even. Sufficient.
(2) \(y=4-x^2\)
If \(x\) is odd, then \(y=even-odd=odd\), but if \(x\) is even, then \(y=even-even=even\). Not sufficient.
Answer: A
General Discussion
shasadou
Joined: 12 Aug 2015
Last visit: 24 Nov 2022
Posts: 219
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Given Kudos: 1,475
Concentration: General Management, Operations
Schools: Johnson '18 (WL) Duke '18 (M) Tuck '18 (D)
GMAT 1: 640 Q40 V37
GMAT 2: 650 Q43 V36
GMAT 3: 600 Q47 V27
GPA: 3.3
WE:Management Consulting (Consulting)
06 Feb 2016, 03:05
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wow this was cool. pluging in 1, 0, 2 helps
