(a)
I’ll use the notation ↓ for subscript and ^ for superscript (exponent) wherever necessary.
Let P(x)=∑rᵢxꜞand Q(x)=(s↓n)xⁿ+∑sᵢxꜞfor i ∊[0,n-1].
Therefore:
P(x).(2x+5)+Q(x).(3)=
2x∑rᵢxꜞ+5∑rᵢxꜞ+3∑sᵢxꜞ+3(s↓n)xⁿ=
2∑rᵢxꜞ⁺¹+5∑rᵢxꜞ+3∑sᵢxꜞ+3(s↓n)xⁿ.
In other words Q is one degree higher than P.
From this for xʲ we have the coefficient 2(r↓(j-1))+5(r↓j)+3(s↓j) for j ∊[1,n-1]. For example, if j=1, we have 2r₀x+5r₁x+3s₁x.
Now we need to look at what happens for the highest degree and the lowest (constant term).
At the highest degree xⁿ, the coefficient is 2(r↓(n-1))+3(s↓n).
The constant term is 5r₀+3s₀=1 because the other x terms are variable and cannot therefore sum to 1 for general x. In fact, all terms involving x must sum to zero.
Let’s take an example to make this clearer: P(x)=r₂x²+r₁x+r₀, Q(x)=s₃x³+s₂x²+s₁x+s₀.
(2x+5)P+3Q=
(2x+5)(r₂x²+r₁x+r₀)+3(s₃x³+s₂x²+s₁x+s₀)=
2r₂x³+2r₁x²+2r₀x
+5r₂x²+5r₁x+5r₀+
3s₃x³+3s₂x²+3s₁x+3s₀
Thus: x³(2r₂+3s₃)+x²(2r₁+5r₂+3s₂)+x(2r₀+5r₁+3s₁)+(5r₀+3s₀).
The highest and lowest terms have a different pattern of coefficients from those in between. In this example, we conclude:
2r₂+3s₃=0,
2r₁+5r₂+3s₂=0,
2r₀+5r₁+3s₁=0,
5r₀+3s₀=1.
We can write for the constant terms: r₀=⅕(1-3s₀). If s₀=2, r₀=-1, and there is a series of pairs of integer values (s₀,r₀): (2,-1), (7,-4), (12,-7),... (-3,2), (-8,5), (-13,8),... Call S the set of such pairs. Call Sᵣ={... -7 -4 -1 2 5 ...}, the set of r₀ within the pairs in S. For any integer t₀, S={(2-5t₀,3t₀-1)}. t₀ is a parameter linking s₀ and r₀. Any integer t₀ will produce a valid (s₀,r₀) pair in S.
2r₀+5r₁+3s₁=0, so 2(3t₀-1)+5r₁+3s₁=0, 6t₀-2+5r₁+3s₁=0, 3(2t₀+s₁)+5r₁=2, r₁=⅕(2-3(2t₀+s₁)). When 2t₀+s₁=5t₁-1, where t₁ is an independent integer, r₁=1-3t₁, s₁=5t₁-2t₀-1.
2r₁+5r₂+3s₂=0, so 2(1-3t₁)+5r₂+3s₂=0, 2-6t₁+5r₂+3s₂=0, 3(-2t₁+s₂)+5r₂=-2, r₂=⅕(-2-3(-2t₁+s₂)). When -2t₁+s₂=5t₂+1, where t₂ is an independent integer, r₂=-1-3t₂, s₂=5t₂+2t₁+1.
So far we have generated integers s₀, r₀, s₁, r₁, s₂, r₂ from arbitrary integers t₀, t₁, t₂.
In the example, the last stage is 2r₂+3s₃=0. Substituting for r₂:
2(-1-3t₂)+3s₃=0, s₃=⅔(1+3t₂)=⅔+2t₂. So s₃ cannot be an integer. This seems to prove that 2x+5 and 3 cannot be combined with any second degree P(x) and third degree Q(x). But we need to extend the argument generally for any degree.
Whatever the value of n, the first stage (determining s₀ and r₀ in terms of a parameter t₀) is the same. The last stage is 2(r↓(n-1))+3(s↓n)=0, so s↓n=-⅔(r↓(n-1)). The intermediate stages are:
2(r↓(i-1))+5(r↓i)+3(s↓i)=0 for i ∊ [1,n-1].
Let’s recap how r in parameter form is expressed at each level:
r₀=3t₀-1, r₁=1-3t₁, r₂=-1-3t₂.
Taking this to the next level, to find r₃ we use 2r₂+5r₃+3s₃=0 using i=3 in the above formula.
2(-1-3t₂)+5r₃+3s₃=0, -2-6t₂+5r₃+3s₃=0, r₃=⅕(2+6t₂-3s₃)=⅕(2+3(2t₂-s₃)). If 2t₂-s₃=5t₃+1, r₃=⅕(2+15t₃+3)=1+3t₃.
The conclusion is that each r is 3T±1 where T is any integer. So at the last stage s↓n=-⅔(r↓(n-1))=-⅔(3T±1)=2T±⅔, which is not an integer, thus proving that 2x+5 and 3 can’t be combined.
(b)
Let P(x)=r₁x+r₀ and Q(x)=s₂x²+s₁x+s₀, then:
(r₁x+r₀)(2x+5)+4(s₂x²+s₁x+s₀)=1.
2r₁x²+5r₁x
+2r₀x+5r₀+
4s₂x²+4s₁x+4s₀=1.
(1) 2r₁+4s₂=0
(2) 2r₀+5r₁+4s₁=0
(3) 5r₀+4s₀=1
From (3) we can see straight away that r₀=1 and s₀=-1 satisfies the equation. But other values are possible, for example, r₀=-3, s₀=4.
Substitute r₀=1 in (2):
2+5r₁+4s₁=0, r₁=⅕(-4s₁-2). When s₁=2, r₁=-2.
In (1), substitute r₁=-2: -4+4s₂=0, so s₂=1.
So, P(x)=-2x+1, Q(x)=x²+2x-1.
CHECK
(-2x+1)(2x+5)+4(x²+2x-1)=-4x²-8x+5+4x²+8x-4=1.
(c)
25 gives P=10x-11 and Q=-6x²+3x+4. 20 gives no valid polynomials, There’s insufficient space to continue this answer.