Let \(J\) be the ideal in \(\mathbb Z[x]\) generated by \(x-5\) and \(14\). Find \(n\in\mathbb Z\) such that \(0\le n\le13\) and \[ (x^3+2x+1)^{50}-n\in J. \]
Solution
1. For \(f\in\ZZ[x]\) and \(m\in\ZZ\), one has \(f-m\in J\) if and only if\(f(5)\equiv m\pmod{14}\).
Because \(J=(x-5,14)\), the quotient satisfies \[ \ZZ[x]/J \cong (\ZZ/14\ZZ)[x]/(x-5) \cong \ZZ/14\ZZ, \] and under this isomorphism the class of \(f\) is the residue class of \(f(5)\) modulo \(14\).
2. One has\[ 136^{50}\equiv2\pmod{14}. \]
Since \[ 136\equiv10\pmod{14}, \] it suffices to compute \(10^{50}\) modulo \(14\). Modulo \(2\), \[ 10^{50}\equiv0\pmod2. \] Modulo \(7\), \[ 10^{50}\equiv3^{50}. \] Since \(3^6\equiv1\pmod7\) and \(50\equiv2\pmod6\), \[ 3^{50}\equiv3^2\equiv9\equiv2\pmod7. \] By the Chinese remainder theorem, the unique residue modulo \(14\) that is \(0\) modulo \(2\) and \(2\) modulo \(7\) is \(2\).
3. The required integer is\[ \boxed{n=2}. \]
2. Step 1 gives\[ (x^3+2x+1)^{50}-2\in J, \]and \(0\le2\le13\).
4. Q.E.D.
Step 3 gives the requested \(n\).