Does this Turing Machine decidelanguage of polynomials with integer coefficients which have integer roots:The input represents a polynomial over variables x1, . . . , xn with integer coefficients.1. Examine all possible integer values of x1, . . . , xn.2. Evaluate the polynomial on all of them.3. If any of them evaluates to 0, accept; else reject  From what I know I would say No, since: M decides a language L iff M accepts all strings in L and reject all strings not in L . Where L is the language of polynomials.  Is this correct?Any help is appreciated!
Does this Turing Machine M decide the language of polynomials?
computability;turing machines
The problem with this algorithm is that it never terminates when the polynomial has no integer roots. Its semantics are:If the input polynomial is in $L$, the machine halts and answers YES.If the input polynomial is not in $L$, the machine never halts.The semantics we are after are a bit different:If the input polynomial is in $L$, the machine halts and answers YES.If the input polynomial is not in $L$, the machine halts and answers NO.