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If m, n and p are constants, m < n < p, and y^3 - y = (y + m)(y + n)(y

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Bunuel
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If m, n and p are constants, m < n < p, and y^3 - y = (y + m)(y + n)(y [#permalink] New post   07 Dec 2020, 09:30
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If m, n and p are constants, \(m < n < p\), and \(y^3 - y = (y + m)(y + n)(y + p)\) for all numbers y, what is the value of n?

A. -3
B. -1
C. 0
D. 1
E. 3
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Re: If m, n and p are constants, m < n < p, and y^3 - y = (y + m)(y + n)(y [#permalink] New post   07 Dec 2020, 09:40
Bunuel wrote:
If m, n and p are constants, \(m < n < p\), and y^3 - y = (y + m)(y + n)(y + p) for all numbers y, what is the value of n?

A. -3
B. -1
C. 0
D. 1
E. 3


LHS becomes \(y(y^2 - 1) = y(y-1)(y+1) \)

Thus \(y(y-1)(y+1) = (y + m)(y + n)(y + p)\)

Since \(m < n < p\) Thus \(Y = Y + N\) and since this is only possible if N = 0. Ans C.
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Re: If m, n and p are constants, m < n < p, and y^3 - y = (y + m)(y + n)(y [#permalink] New post   07 Dec 2020, 09:53
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Asked: If m, n and p are constants, \(m < n < p\), and y^3 - y = (y + m)(y + n)(y + p) for all numbers y, what is the value of n?

y^3 - y = = y(y^2 - 1) = (y-1)y(y+1) = (y + m)(y + n)(y + p)
m = -1; n=0; p = 1

IMO C
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Re: If m, n and p are constants, m < n < p, and y^3 - y = (y + m)(y + n)(y [#permalink] New post   07 Dec 2020, 10:33
Bunuel wrote:
If m, n and p are constants, \(m < n < p\), and \(y^3 - y = (y + m)(y + n)(y + p)\) for all numbers y, what is the value of n?

A. -3
B. -1
C. 0
D. 1
E. 3

\(y^3 - y = (y + m)(y + n)(y + p)\)
\((y-1)y(y+1) = (y + m)(y + n)(y + p)\)
\(\implies\) for y integer (y + m)(y + n)(y + p) are consecutive integers
However, more importantly, whether
(y+m) < (y+p) for all y,m and p either positive or negative OR y(+), m(-) and p(-)
OR
(y+m) > (y+p) for all y,m and p either negative or y(+) and m(-) and p(-).

y+n = y [(y+n) will always be in between (y+m) and (y+p) without caring about whether y, m, n and p are integers or not]
\(\implies\) n = 0

Answer C.
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Re: If m, n and p are constants, m < n < p, and y^3 - y = (y + m)(y + n)(y [#permalink]
07 Dec 2020, 10:33
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