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Let $a, b, c$ be roots of the polynomial $x^3-3x^2-4x+5$. Compute the remainder when $[\frac{a^4+b^4}{a+b} + \frac{b^4+c^4}{b+c} + \frac{c^4+a^4}{c+a}]$ is divided by $7$, where $[x]$ denotes floor function.

Well, I tried finding out the value of the summation of the fourth powers and then substituting them in the individual terms.

I got something like this $[\frac{a^4+b^4}{a+b} + \frac{b^4+c^4}{b+c} + \frac{c^4+a^4}{c+a}]$ = $[\frac{197 - c^4}{3-c} + \frac{197 - a^4}{3-a} + \frac{197 - b^4}{3+b}]$

I am not sure how to proceed after this, as by taking the L.C.M of the denominators and converting it into an expanded form would result in a long fraction. Is there any more efficient way to solve this?

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    $\begingroup$ Just out of curiosity : how much time is given for this problem ? $\endgroup$ Commented Aug 4, 2023 at 7:44
  • $\begingroup$ it is a five marker from a mock test for the ioqm exam held in India , i guess five markers take about 15-20 mins $\endgroup$ Commented Aug 4, 2023 at 7:45
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    $\begingroup$ How about something like: $\frac{197-c^4}{3-c} = c^3 + 3c^2 + 9c + 27 - \frac{116}{c-3}$ and similarly for the others. And then, to find $\sum_{cyc} \frac{1}{a-3}$, form the polynomial with $a-3,b-3,c-3$ with roots, and then the polynomial with $(a-3)^{-1}, (b-3)^{-1}, (c-3)^{-1}$ as roots. $\endgroup$ Commented Aug 4, 2023 at 17:15
  • $\begingroup$ @DanielSchepler good idea .. $\endgroup$ Commented Aug 4, 2023 at 17:37
  • $\begingroup$ Also, note the fractions divide out. $\frac{a^4+b^4}{a+b} = -a^2b +ab^2-b^2+bc^2-c^2a+ca^2> The resulting 6 terms factor a bit. I started the exercise then looked at the answers to see if I was going the right direction. Upon seeing a much simpler answer I just stopped. $\endgroup$
    – nickalh
    Commented Aug 5, 2023 at 1:03

5 Answers 5

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I would like to point that there [was] an error in the accepted answer since I can not comment. It has been wrongfully evaluated that $(a+b)(b+c)(c+a)=2/3$.

Rather, we can use the well known identity: $$(a+b+c)(ab+bc+ca)=(a+b)(b+c)(c+a)+abc$$ to evaluate that $(a+b)(b+c)(c+a)=-7$.

Hence, the expression is $869/7$ as also pointed in the comments. $$\lfloor\frac{869}{7}\rfloor=124$$ hence $$124 \equiv \fbox{5} \pmod 7$$

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Consider the polynomial $x^3 - 3x^2 - 4x + 5 = 0$. It has been given that $a, b, c$ are the roots of this polynomial.

Thus we have $$ a^4 + b^4 - 3a^3 - 3b^3 - 4a^2 - 4b^2 + 5a + 5b = 0 $$

Hence, $$ a^4 + b^4 = 3a^3 + 3b^3 + 4a^2 + 4b^2 - 5a - 5b$$

Now, dividing by $a + b$

$$ \frac{a^4 + b^4}{a+b} = \frac{3a^3 + 3b^3 + 4a^2 + 4b^2 - 5a - 5b}{a+b}$$

$$\frac{a^2 + b^2}{a+b} = (a+b) - \frac{2ab}{a+b}$$ $$\frac{a^3 + b^3}{a+b} = \frac{(a+b)^3 - 3ab(a+b)}{a+b} = a^2 + b^2 + 2ab - 3ab = a^2 + b^2 - ab$$

$$ \frac{3a^3 + 3b^3 + 4a^2 + 4b^2 - 5a - 5b}{a+b} = 3\frac{a^3 + b^3}{a+b} + 4\frac{a^2 + b^2}{a + b} - 5 = 3(a^2 + b^2) - 3ab + 4a + 4b - 8ab/(a+b) - 5$$

On adding all three, we get $$ 6(a^2 + b^2 + c^2) -3(ab+bc+ca) + 8(a+b+c) -8\left(\frac{(ab+bc+ca)^2 + abc(a+b+c)}{(a+b)(b+c)(c+a)}\right) - 15 $$

There remain the following set of equations,

$$6(a^2 + b^2 + c^2) = 6((a+b+c)^2 - 2(ab + bc + ca)) = 6(9 + 2(4)) = 6 \times 17 = 102 $$ $$-3(ab+bc+ca) = 12$$ $$8(a+b+c) = 8\times3 = 24$$ $$(ab+bc+ca)^2 = 16$$ $$abc(a+b+c) = -5\times 3 = -15$$

Thus we have $$ 102 + 12 + 24 - 15 - \frac{8}{(a+b)(b+c)(c+a)} = 123 - \frac{8}{(a+b)(b+c)(c+a)}$$

Edit By using the given identity (courtsey of user @Sahaj) $$(a+b)(b+c)(c+a) = (a+b+c)(ab+bc+ca) - abc = 3\times-4 + 5 = -7$$

Hence the expression finally becomes $$123 + 8/7 = \frac{869}{7}$$, whose greatest integer is $124$

Hence the remainder is $5$.

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    $\begingroup$ Well that was a monster of an answer , i never expected it to be such big , but thanks for the efforts you made . Bravo !! $\endgroup$ Commented Aug 4, 2023 at 13:32
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    $\begingroup$ Hmm, when I put the problem through Sagemath, I got the expression being equal to $\frac{869}{7}$. So, the floor would be 124, which has remainder 5 when divded by 7. $\endgroup$ Commented Aug 4, 2023 at 17:03
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    $\begingroup$ @SwagBeast SkJJ , it would be nice , if you edit your answer according to the calculations of Sahaj $\endgroup$ Commented Aug 5, 2023 at 3:59
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    $\begingroup$ I figured my error, thanks @Sahaj $\endgroup$ Commented Aug 5, 2023 at 11:03
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Another solution. Continuing from $$\sum \frac{197-c^4}{3-c} = \sum \frac{116}{3-c}+\sum \frac{81-c^4}{3-c}$$ $$=116\cdot \frac{P'(3)}{P(3)}+\sum(3+c)(9+c^2)$$ $$=116\cdot \frac{5}{-7}+\sum(27+3c^2+9c+c^3)$$ $$=-\frac{580}{7}+81+3\cdot17+9\cdot3+48$$ $$=207-\frac{580}{7}=\frac{869}{7}$$

Remark : I noticed that @DanielSchelper already pointed out this approach, but the result I used in second line makes this a quicker method : $$\frac{P'(x)}{P(x)}=\sum_{P(\alpha)=0} \frac{1}{x-\alpha}$$

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As already observed, $$ \sum_{cyc}\frac{a^4+b^4}{a+b}=\sum_{cyc}\frac{s_4-a^4}{s_1-a} $$ where $s_k=\sum_{cyc}a^k$ are the power sums over the roots.

The cubic polynomial $p(x)=x^3+c_2x^2+c_1x+c_0$ can be decomposed into $$ p(x)=(x+c_2)(x^2+c_1)+c_0-c_1c_2 $$

Note that $c_2=-s_1$ by the Viete identities. Thus the expression in question can be further transformed to $$ \sum_{cyc}\frac{a^4+b^4}{a+b} =\sum_{cyc}\frac{a^4-s_4}{a+c_2} =\sum_{cyc}\frac{(a^4-s_4)(a^2+c_1)}{c_1c_2-c_0} \\~\\ =\frac{s_6-s_4s_2+s_4c_1-3s_4c_1}{c_1c_2-c_0} =\frac{s_6-(s_2+2c_1)s_4}{c_1c_2-c_0} =\frac{s_6-s_1^2s_4}{c_1c_2-c_0} $$ using the binomial or Newton identity $s_1^2=s_2+2c_1$.

Inserting that the power sums are $\{s_k\}_k=(3, 3, 17, 48, 197, 698, 2642, 9733, 36277, 134553, 500102, 1857133, 6899042,...)$ gives $$ \sum_{cyc}\frac{a^4+b^4}{a+b}=\frac{2642-3^2\cdot 197}{7}=\frac{869}{7}=124+\frac17 $$

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$$\sum_{cyc}\frac{a^4+b^4}{a+b}=\frac ND=\frac{869}{7}$$ where $$N=\sum_{cyc}(p_4-c^4)(p_1-a)(p_1-b)=e_2p_4+p_1^2p_4-p_1p_5-e_3p_3=-869$$

$$D=\prod_{cyc}(p_1-a)=27-9p_1+3e_2-e_3=-7$$ by using the values $e_1=p_1=3$, $e_2=-4$, $e_3=-5$, $p_2=17$, $p_3=48$, $p_4=197$, $p_5=689$ found by Vieta formulae and Newton-Girard identities.

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