Let \(I\subset\mathbb Z[x]\) be the ideal generated by \(13\) and \(x-4\). Find an integer \(m\) with \[ 0\le m\le12 \] such that \[ (x^{26}+x+1)^{73}-m\in I. \]
Solution
1. In the quotient ring\[ \ZZ[x]/I, \]one has\[ 13=0 \qquad\text{and}\qquad x=4. \]
The ideal \[ I=(13,x-4) \] contains both \(13\) and \(x-4\). Thus their residue classes vanish, which is exactly the two displayed relations.
2. Modulo \(I\),\[ x^{26}+x+1 \equiv 8. \]
By step 1, \[ x^{26}+x+1 \equiv 4^{26}+4+1 \pmod I. \] Working modulo \(13\), Fermat’s theorem gives \[ 4^{12}\equiv1\pmod{13}. \] Since \[ 26=2\cdot12+2, \] one gets \[ 4^{26} \equiv 4^2 =16 \equiv3 \pmod{13}. \] Therefore \[ 4^{26}+4+1 \equiv 3+4+1 =8 \pmod{13}, \] which gives the congruence modulo \(I\).
3. Modulo \(13\),\[ 8^{73}\equiv8. \]
Again by Fermat’s theorem, \[ 8^{12}\equiv1\pmod{13}. \] Since \[ 73=6\cdot12+1, \] one has \[ 8^{73} \equiv 8 \pmod{13}. \]
4. Modulo \(I\),\[ (x^{26}+x+1)^{73} \equiv 8. \]
Step 2 gives \[ (x^{26}+x+1)^{73} \equiv 8^{73} \pmod I. \] The integer relation from step 3 is also valid modulo \(I\) because \(13\in I\). Thus the right side is congruent to \(8\).
5. One may take\[ \boxed{m=8}. \]
The integer \(8\) lies in the required range \[ 0\leq8\leq12. \] Step 4 says exactly that \[ (x^{26}+x+1)^{73}-8\in I. \]
6. Q.E.D.
Step 5 gives the required integer.