The equation $$ a_n x^n+\dots+a_0=1, \qquad a_n\neq 0, \qquad a_i\in\mathbb{Z} $$ has $\geq 4$ distinct integral solutions. How can I prove that $$ a_n x^n+\dots+a_0=-1 $$ has no integral solution?
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It does have an integral solution. Take x = 0 and a_0 = -1. – D.B. Nov 23 '18 at 18:47
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2@D.B. but since $a_i$ are integral, you can not satisfy the first condition with $a_0=-1$ – Nov 23 '18 at 18:48
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Hint:
Let $f(x)=\sum_{k=0}^na_kx^k$. Suppose $p,q,r,s$ are four distinct integers such that $$f(p)=f(q)=f(r)=f(s)=1.$$ Then, $$f(x)=(x-p)(x-q)(x-r)(x-s) \,\,g(x)+1, \quad \text{where } g(x) \text{ is some polynomial in } \Bbb{Z}[x].$$ If $t \in \Bbb{Z}$ is such that $f(t)=-1$. Then, $$f(t)=-1=(t-p)(t-q)(t-r)(t-s) \,\,g(t)+1.$$ This is same as saying $$(t-p)(t-q)(t-r)(t-s) \,\,g(t)=-2$$ Observe that $t \not\in\{p,q,r,s\}$ and the left side is a product of integers. Can you now arrive at a contradiction?
Anurag A
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