Division theorem for polynomials with integer coefficients I can see that the Division Theorem holds for polynomials in $\mathbb{Q}[x]$, but does not necessarily hold for polynomials in  $\mathbb{Z}[x]$, e.g.

Let $f=x^2+3x$ and $g=5x+2$.

Then the Division Theorem yields unique polynomials $q$ and $r$:

$f=gq+r$, in essence,

$x^2+3x=(\frac{1}{5}x+\frac{13}{25})\cdot(5x+2)-\frac{26}{25}$

where $q=\frac{1}{5}x+\frac{13}{25}$ and $r=-\frac{26}{25}$, but $q,r \notin \mathbb{Z}[x]$.


  The question is, how does one show that any $q,r$ fails to be in $\mathbb{Z}[x]$ to satisfy $f=gq+r$? 


More concretely, for the existence of some $f$ and $g$ in $\mathbb{Z}[x]$, how does one show that there does not exist $q,r$ in $\mathbb{Z}[x]$ such that $f=gq+r$ with the property that $r=0$ or $\deg(r)&lt;\deg(g)$?
If the leading coefficient of $g$ does not divide the leading coefficient of $f$ (in $\mathbb Z$) then there are no $q,r\in\mathbb Z[X]$ such that $f=gq+r$ with $r=0$ or $\deg r&lt;\deg g$.
It's kind of silly honestly, but if you can't divide f by g for some reason like if g has a higher degree what we do we just set q to 0 and r to f and the statement holds.

but that's not super satisfying so what about if the degree of g is less than or equal well I can come up with something that still works. So let's refine what we mean by f. I'm gonna say f is a sum of variables and coefficients of this form 

f = (a0)+(a1)x+(a2)x^2+...(an)*x^n (i don't know how to add subscripts)

basically I'm saying f is a polynomial and I'm listing out the x's and their coefficients because I'm gonna use the big (an)*x^n soon. I'm also gonna do the same thing for g. so g= (b0)+(b1)x+(b2)x^2+...(bk)*x^k. 

so I'm gonna make a q now for f.
so I wanna call f1 = f -[a(n-k)/b(k)]  *x^(n-k)*g 

: note (a(n-k)) is a coefficient in g not a* (n-k) again I don't know how to add subscripts yet

this is where I think the crux of your problem is, we need the biggest coefficient in f do be divisible by the biggest coefficient in g if we don't have it, then it doesn't divide any further we're stuck. Luckily that theorem assumes that (bk) is a unit

here's a quoted definition
"Let R be a commutative ring. Let f , g be two polynomials in R[x] with [g] ̸= 0, and suppose that the leading co- efficient of [g] is a unit of R..."

i.e. 

that all this time an has to have a defined inverse in our assumptions.

now this gets easier we're nearly done.

f= f1+ [a(n-k)/b(k)]  *x^(n-k)*g 
Now what?we induct. What do I mean we induct? well we have a base case let degree of f be 0, then if we remember at the top we can just let the whole expression be f=0*g+r where r=f. So now we can induct upwards using complete induction.

so
f= f1+ [a(n-k)/b(k)]*x^(n-k)g 
 = gq1+r1+[a(n-k)/b(k)]*x^(n-k)g
 = [ gq1+[a(n-k)/b(k)]*x^(n-k)*g ]  +r1

= g*(q1+[a(n-k)/b(k)]*x^(n-k)*g)       +r1
hey wait this is what we wanted to prove.
done.