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Prove that there is no polynomial with integer coefficients $P(x)$ such that $P(7)=5$ and $P(15)=9$.

In general; how to know if there exists a polynomial with integer coefficients $P(a)=b$ and $P(c)=d$, where $a,b,c$, and $d$ are all integers?

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For each $a,b\in \mathbb{Z}$ and $P\in \mathbb{Z}[X]$ we have $$a-b\mid P(a)-P(b)$$ so $$15-7\mid P(15)-P(7)= 9-5\implies 8\mid 4$$

A contradiction!

On the other hand if $$a-c\mid b-d$$ then such a polynomial exist. Take $$P(x) ={d-b\over c-a} (x-a)+b$$

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  • $\begingroup$ Beware The proof is incomplete / incorrect since it never used the hypothesis that $P(x)$ has integer coefficients - which is crucial. $\endgroup$ – Bill Dubuque Oct 30 '18 at 14:15
  • $\begingroup$ But this is said in parenthesis. @BillDubuque $\endgroup$ – Maria Mazur Oct 30 '18 at 14:16
  • $\begingroup$ It was never mentioned in your original version. Your edit that repeats the hypothesis does not fix the problem. You need to justify why the claimed divisibility holds true for polynomials with integer coefficients. $\endgroup$ – Bill Dubuque Oct 30 '18 at 14:18
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.A polynomial with integer coefficients $P$ always satisfies the relation that if $a,b$ are integers, then $P(b) - P(a)$ is a multiple of $b-a$. This can be seen by showing that $b^k - a^k$ is a multiple of $b-a$ for all natural $k$, and then noting that $P(b) - P(a)$ is a linear combination of these.

Now, your polynomial does not satisfy this property : $ 9-5$ is not a multiple of $15-7$. Hence it does not have integer coefficients.

On the other hand, at least in the two variable case, if we have $f(a) = b,f(c) = d$ and $a-c$ divides $b-d$, then an answer with integral coefficients can be found by taking the polynomial whose graph is a straight line passing through $(a,b)$ and $(c,d)$.

In the three variable, and multivariable case, you may want to read up the Lagrange interpolating polynomial. Look at the coefficients of this polynomial, and see what conditions on $x_i, f(x_i)$ can make them integral. Derive conditions for existence of such a polynomial for the multivariable case, and see the case $n=2$ from here. Contradictions on existence of such a polynomial are also shown in the same manner as we did for the $2$ variable case.

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