Let \(I=(7,x-3)\) in \(\mathbb Z[x]\).
- Prove that for every \(r\in\mathbb Z[x]\) there is an integer \(\alpha\) with \(0\le\alpha\le6\) such that \[ r-\alpha\in I. \]
- Find \(\alpha\) for \[ r=x^{250}+15x^{14}+x^2+5. \]
Solution
1. For every \(r\in\ZZ[x]\),\[ r(x)-r(3)\in(x-3). \]
By the factor theorem, the polynomial \(r(x)-r(3)\) vanishes at \(x=3\), so it is divisible by \(x-3\) in \(\ZZ[x]\).
2. For every \(r\in\ZZ[x]\), there is a unique integer\(\alpha\) with \(0\leq\alpha\leq6\) such that\[ r(3)-\alpha\in(7). \]
The integers \(0,1,\ldots,6\) form a complete set of representatives for the residue classes modulo \(7\). Choose \(\alpha\) to represent the class of the integer \(r(3)\).
3. The integer \(\alpha\) from step 2 satisfies\[ r-\alpha\in I. \]
We can write \[ r-\alpha = \bigl(r-r(3)\bigr) + \bigl(r(3)-\alpha\bigr). \] By step 1, the first summand lies in \((x-3)\), and by step 2 the second lies in \((7)\). Hence their sum lies in \[ (7,x-3)=I. \] This proves part 1.
4. For\[ r=x^{250}+15x^{14}+x^2+5, \]one has\[ r(3)\equiv6\pmod7. \]
Since \[ 3^6\equiv1\pmod7, \] we have \[ 3^{250}\equiv3^4\equiv4\pmod7 \] and \[ 3^{14}\equiv3^2\equiv2\pmod7. \] Also \(15\equiv1\pmod7\). Therefore \[ \begin{aligned} r(3) &= 3^{250}+15\cdot3^{14}+3^2+5\\ &\equiv 4+2+2+5\\ &\equiv 6 \pmod7. \end{aligned} \]
5. For the polynomial in part 2,\[ \boxed{\alpha=6}. \]
Step 4 shows that the unique representative of \(r(3)\) modulo \(7\) in the range \(0\leq\alpha\leq6\) is \(6\).
6. Q.E.D.
Step 3 proves part 1, and step 5 proves part 2.