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What is wrong is the following argument ?

The degree of $phi-phi'$ is d (4 in your example), which is smaller than the degree of h (which is 8 in your example). Hence, $phi-phi' = 0 mod h$ is equivelent to $phi-phi' = 0$. Hence $phi = 0'$ is the only solution to the equation.

What is wrong is the following argument ?

The degree of $phi-phi'$ is d (4 in your example), d, which is smaller than the degree of h (which is 8 in your example). Hence, $phi-phi' = 0 mod h$ is equivelent to $phi-phi' = 0$. Hence $phi = 0'$ is the only solution to the equation.since $d<h$. hence,="" $phi-phi'="0" mod="" h$="" is="" equivalent="" to="" $phi-phi'="0$." hence="" $phi="0'$" is="" the="" only="" solution="" to="" the="" equation.<="" p="">

What is wrong is the following argument ?

The degree of $phi-phi'$ is d, which is smaller than the degree of h since $d<h$. hence,="" $phi-phi'="0" mod="" h$="" is="" equivalent="" to="" $phi-phi'="0$." hence="" $phi="0'$" is="" the="" only="" solution="" to="" the="" equation.<="" p="">

What is wrong is the following argument ?

The degree of $phi-phi'$ $\phi-\phi'$ is d, which is smaller than the degree of h since $d<h$. hence,="" $phi-phi'="0"$\phi-\phi'="0" mod="" h$="" is="" equivalent="" to=""$phi-phi'="0$."$\phi-\phi'="0$." hence=""$phi="0'$"$\phi="0'$" is="" the="" only="" solution="" to="" the="" equation.<="" p=""> What is wrong is the following argument ? The degree of$\phi-\phi'$is d, which is smaller than the degree of h since$d<h$. hence,=""$\phi-\phi'="0" mod="" h$="" is="" equivalent="" to=""$\phi-\phi'="0$." hence=""$\phi="0'$" is="" the="" only="" solution="" to="" the="" equation.<="" p=""> What is wrong is the following argument ? The degree of$\phi-\phi'$is d, which is smaller than the degree of h since$d<h$.$d<h$.="" hence,=""$\phi-\phi'="0" mod="" h$="" is="" equivalent="" to=""$\phi-\phi'="0$." hence=""$\phi="0'$" is="" the="" only="" solution="" to="" the="" equation.<="" p=""> What is wrong is the following argument ? The degree of$\phi-\phi'$is d, which is smaller than the degree of h since$d<h$.="" hence,=""$\phi-\phi'="0" mod="" h$="" is="" equivalent="" to=""$\phi-\phi'="0$." hence=""$\phi="0'$" is="" the="" only="" solution="" to="" the="" equation.<="" .<="" p=""> Hence,$\phi-\phi' = 0 \mod h$is equivalent to$\phi-\phi' = 0$. Hence$\phi = 0'$is the only solution to the equation. What is wrong is the following argument ? The degree of$\phi-\phi'$is d, which is smaller than the degree of h since$d<h$.<="" p=""> Hence,$\phi-\phi' = 0 \mod h$is equivalent to$\phi-\phi' = 0$. Hence$\phi = 0'$0$ is the only solution to the equation.

What is wrong is the following argument ?

The degree of $\phi-\phi'$ is d, which is smaller than the degree of h since $d<h$ .<="" $d<h$.< p="">

Hence, $\phi-\phi' = 0 \mod h$ is equivalent to $\phi-\phi' = 0$. Hence $\phi = 0$ is the only solution to the equation.

What is wrong is the following argument ?

The degree of $\phi-\phi'$ is d, which is smaller than the degree of h since $d<h$.< $d<h$< p="">

Hence, $\phi-\phi' = 0 \mod h$ is equivalent to $\phi-\phi' = 0$. Hence $\phi = 0$ is the only solution to the equation.

What is wrong is the following argument ?

The degree of $\phi-\phi'$ is d, which is smaller than the degree of h since $d<h$< $d<h$ <="" p="">

Hence, $\phi-\phi' = 0 \mod h$ is equivalent to $\phi-\phi' = 0$. Hence $\phi = 0$ is the only solution to the equation.