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The problem states:

Find the remainder when $x^{73}-2x^{15}+3x-1$ is divided by $(x-1)^2$.

So, what I did is,

Assume, the remainder to be linear, i.e $r(x) = ax+b$

By Eucild''s,

$x^{73}-2x^{15}+3x-1 = q(x)(x-1)^2+r(x)$

Putting x = 1,

We get, $a+b = 1$.

Then I differentiated it, and the rest is trivial.

How can I skip differentiation and get the answer by some other method/way. I was thinking perhaps something involving number theory? I want to know is it possible to skip calculus in such questions?

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    $\begingroup$ You could compute the $x$ and $1$ term of $(x + 1)^{73} - 2(x+1)^{15} + 3(x + 1) - 1$ using the binomial theorem. It's not as elegant as the differentiation method, but I fear that nothing else will be. $\endgroup$
    – Theo Bendit
    Commented Mar 20 at 12:07
  • $\begingroup$ I did not quite get it. $\endgroup$
    – Krave37
    Commented Mar 20 at 12:21
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    $\begingroup$ You can avoid differentiation by doing the division and seeing what the remainder is. But it's likely to be (a lot) more work. $\endgroup$
    – Henrik supports the community
    Commented Mar 20 at 14:09
  • $\begingroup$ In general, when dividing $p(x)$ with $(x-a)^2$, you'll get $p'(a)(x-a)+p(a)$ as the remainder. Hence it seems straightforward to find $p'(x)$; as already mentioned in comments, other approaches are cumbersome though possible. To add to the list, you can find the polynomial $q(x)=(p(x)-p(a))/(x-a)$ and get $q(a)=p'(a)$, essentially dancing around differentiation using some algebra. In practise probably hellish to do the polynomial division by hand, I expect all 73 terms to appear in $q(x)$. $\endgroup$
    – Macavity
    Commented Mar 20 at 15:05

4 Answers 4

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Claim: $$x^n\equiv nx-(n-1)\pmod{(x-1)^2}$$ Proof by induction: It is true for $n=2$. Let it be true for $n$ as above. For $n+1$, $$x^n\equiv nx-(n-1)\implies\\ x^{n+1}\equiv nx^2-(n-1)x\equiv n(2x-1)-(n-1)x\equiv (n+1)x-n\pmod{(x-1)^2}$$ Using the claim for the problem, we have $$x^{73}-2x^{15}+3x-1\equiv 73x-72-2(15x-14)+3x-1\\ \equiv 46x-45\pmod{(x-1)^2}$$

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  • $\begingroup$ $x^n \equiv nx-(n-1) \pmod (x-1)^2$ How did you get that? $\endgroup$
    – Krave37
    Commented Mar 20 at 17:35
  • $\begingroup$ @Krave37 There presumably you come back to derivatives or binomial theorem $x^n=(1+\overline{x-1})^n = 1+n(x-1)+(x-1)^2(\dots)$. Still simple approach (+1) considering some of the other non-calculus ones. $\endgroup$
    – Macavity
    Commented Mar 20 at 17:43
  • $\begingroup$ Can you update your answer to include this using binomial? $\endgroup$
    – Krave37
    Commented Mar 20 at 17:45
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This answer is to explain my comment more thoroughly.

You are looking for a polynomial $r(x) = ax + b$ such that

$$x^{73}-2x^{15}+3x-1 = q(x)(x-1)^2+r(x)$$

for some polynomial $q(x)$. If we replace $y = x - 1$, i.e. $x = y + 1$, we get an equivalent problem:

$$(y+1)^{73}-2(y+1)^{15}+3(y+1)-1 = q(y+1)y^2+r(y+1).$$

The $q(y+1)y^2$ contains no constant or $y$ terms, and $r(y + 1) = ay + a + b$ contains no $y^2$ terms or higher. So, we simply need to find the constant and $y$ terms in order to figure out $r$.

We can do this with binomial theorem. In particular, we know $$(y + 1)^n = \sum_{k=0}^n \binom{n}{k} y^k = \sum_{k=0}^n \frac{n!}{k!(n-k)!} y^k.$$ So, the constant terms and $y$ terms can be extracted, simply by taking $k = 0$ and $k = 1$ respectively. We have $\binom{n}{0} = 1$ and $\binom{n}{1} = n$. Ignoring quadratic and higher order terms, the left hand side becomes: $$(1 + 73y + \ldots) - 2(1 + 15y + \ldots) + 3(1 + y) - 1 = 1 + 46y + \ldots$$ Therefore, $r(y + 1) = ay + a + b = 1 + 46y$, which implies $a = 46$, and $b = -45$, i.e. $r(x) = 46x - 45$.

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  • $\begingroup$ Actually, this is also brilliant $\endgroup$
    – Krave37
    Commented Mar 21 at 3:45
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I think that your method is completely algebraic and very nice. We just have $a=f'(1)=46$, and hence by $a+b=1$ then $r(x)=46x-45$.

And taking $f'(1)$ is not really "calculus". Differentiation of polynomials in abstract algebra is a linear map $D$ with $D(x^n)=nx^{n-1}$ on monomials. So $f'(1)=D(f)(1)=46$. This is how it is done in abstract algebra. See for example the following posts:

Multiple root of a polynomial and formal derivative.

How does one determine whether a polynomial has a double root?

Repeated roots of a polynomial

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COMMENT.- We know $$P(x)=x^{73}-2x^{15}+3x-1=q_1(x)(x-1)+P(1)=q_1(x)(x-1)+1$$ and because of $q_1(x)=q_2(x)(x-1)+q_1(1)$ we have $$P(x)=q_2(x)(x-1)^2+\color{red}{q_1(1)(x-1)+1}$$ If $P(x)=q_2(x)(x-1)^2+ax+b$ then $P(1)=a+b=1$ and the other equation to determine the values of $a$ and $b$ are given by $q_1(1)$ and $-(q_1(1)-1)$ respectively. The value $q_1(1)$ can be obtained using the binomial theorem and it is $46$ (look at Theo Bendit's answer above).

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