Problem 4

Let

\mathbb{Z}

denote the set of all integers. Find all polynomials

P(x)

with integer coefficients that satisfy the following property: for any infinite sequence

a_1,a_2,\ldots

of integers in which each integer in

\mathbb{Z}

appears exactly once, there exist indices

i<j

and an integer

k

such that

a_i+a_{i+1}+\cdots+a_j=P(k)

Step 5 of 5: Higher odd degree: gaps grow, so a bijection avoiding P can be built

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\deg P\ge3,\ \deg P\text{ odd}:\ \forall A,B,C\ \exists y,\ |y|>C,\ \operatorname{range}(P)\cap[y-A,y+B]=\varnothing

Detailed analysis

If $\deg P$ is odd and at least $3$ , the gap $P(x+1)-P(x)$ grows without bound as $|x|\to\infty$ , so for any bounds $A,B,C$ there is $y$ with $|y|>C$ such that no value of $P$ falls in $[y-A,y+B]$ . Build the sequence greedily: having placed $a_1,\ldots,a_i$ , let $m$ be the unused integer of smallest absolute value, set $a_{i+2}=m$ , and use the gap fact to choose $a_{i+1}$ (all new block sums involve $a_{i+1}$ and lie in a bounded window around it) so that every new consecutive block sum avoids the range of $P$ . This yields a bijection $\mathbb{Z}\to\mathbb{Z}$ with no block sum equal to any $P(k)$ , so $P$ fails the property; together with the previous step, only linear $P$ (with $c\ne0$ ) can satisfy it.