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If the equation $x^n+a_{1}x^{n-1}+a_{2}x^{n-2}+\cdots \cdots +a_{n} = 5$ with integer coefficients has $4$

distinct integer roots . Then total number of integer roots of the equation

$x^{n}+a_{1}x^{n-1}+a_{2}x^{n-2}+\cdots \cdots +a_{n}=7$ is equal to

I did not understand how can i solve it, Help required, Thanks

1 Answers 1

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If $P$ is a polynomial with integer coefficients and $4$ distinct integer roots then $P(n)\neq -2$ for every integer $n$.

To see this write $P$ as $(x-a_1)(x-a_2)(x-a_3)(x-a_4)Q$ such that $Q$ is an integer polynomial.

Notice that $(n-a_1)(n-a_2)(n-a_3)(n-a_4)Q$ is clearly not a prime (because at most two of the factors on the left can be equal to $1$ or $-1$) and hence not equal to $2$.

Conclusion: the answer is zero.

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    To Jorge Fernández Hidalgo how you conclude $P(n)\neq -2,$ – 2017-01-25
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    I explain why later on, in fact it cannot be equal to any prime. – 2017-01-25
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    @juantheron did you read the rest? I showed that $P(n)$ is composite for every $n$. – 2017-01-25
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    Yes Jorge, but i did not understand that line , plz explain me , Thanks – 2017-01-25