Let #P(x)=x^n+5x^(n-1)+3# where #n > 1# is an integer. Prove that #P(x)# cannot be expressed as the product of two polynomials, each of which has all its coefficients integers and degree at least #1#?

2 Answers
George C.
Dec 31, 2017

See explanation...

Explanation:

#P(x) = x^n+5x^(n-1)+3#

Suppose #P(x) = Q(x)R(x)# where:

#Q(x) = a_hx^h+a_(h-1)x^(h-1)+...+a_1x+a_0#

#R(x) = b_kx^k+b_(k-1)x^(k-1)+...+b_1x+b_0#

#h + k = n#

#a_h = b_k = 1#

#a_0 = 3#

#b_0 = 1#

Note that the signs on the constants must be positive, since otherwise #Q(x)# and #R(x)# would have real positive zeros, which #P(x)# does not.

The coefficient of #x^m# in #Q(x)R(x)# is:

#a_mb_0+a_(m-1)b_1+...+a_1b_(m-1)+a_0b_m#

Now let #m# be the least integer such that #a_m# is not a multiple of #3#.

Then modulo #3# the coefficient of #x^m# in #Q(x)R(x)# is:

#a_mb_0 = a_m#

since all of the other terms are multiples of #3#.

Note however that the smallest non-zero coefficient of #P(x)# beyond the constant is that of #x^(n-1)#.

Hence #m >= n-1#, so #Q(x)# is of degree at least #n-1# and #R(x)# is of degree at most #1#.

So either #R(x) = x+1#, which would make #-1# a zero of #P(x)# - which it is not, or #R(x) = 1#, meaning that #P(x)# has no non-trivial polynomial factors with integer coefficients.

Cesareo R.
Dec 31, 2017

See below.

Explanation:

We will show a general procedure but applied to a case study for #n = 9#

Now considering

#P_n(x) = x^n +5x^(n-1)+3 = Q_u(x)R_v(x)#

with

#P_n(x) = sum_(k=0)^n a_k x^k#
#Q_u(x) = sum_(k=0)^u b_k x^k#
#R_v(x) = sum_(k=0)^v c_k x^k#

then

#a_k = sum_(i=0)^k b_(k-i)c_i# (convolution of coefficients)

or exemplifying for #n = 9, u = 5, v = 4#

#{(), (a_0=b_0c_0=3), (a_1=b_1c_0+b_0c_1=0), (a_2=b_2c_0+b_1c_1+b_0c_2=0),(a_3=b_3c_0+b_2c_1+b_1c_2+b_0c_3=0), (cdots), (a_7=b_5c_2+b_4c_3+b_3c_4 = 0), (a_8=b_5c_3+b_4c_4=5),(a_9=b_5 c_5 = 1):}#

Now assuming #b_0 = 3, c_0 = 1# (the other possibility is #b_0 = 1, c_0 = 3#) also #b_5=c_4=1# and #{b_k, c_k} in ZZ#

As we can observe, from the first equation, as a consequence in the second equation

#b_1c_0 equiv b_0 c_1 equiv 0 mod 3 rArr b_1c_1 equiv 0 mod 3# having #b_(k-i)c_i equiv 0 mod 3#

This behavior propagates until

#b_5c_3+b_4c_4=5#

with #b_5b_4 equiv c_3 c_4 equiv 0 mod 3# from the previous equation

As a consequence #b_5c_3+b_4c_4=5# cannot be attained in integers.

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