Divisibility

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Source: Baltic Way 2016, Problem 4

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2105 posts

#1 h Nov 5, 2016, 9:54 PM • 3 Y

Y by Amir Hossein, Adventure10, Mango247

Let $n$ be a positive integer and let $a, b, c, d$ be integers such that $n | a + b + c + d$ and $n | a^2 + b^2 + c^2 + d^2.$
Show that $n | a^4 + b^4 + c^4 + d^4 + 4abcd.$

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#3 h Nov 5, 2016, 10:14 PM • 3 Y

Y by Amir Hossein, ThIsMe, Adventure10

Let $a_i$ be the $i$ -th symmetric sum and let $b_i=a^i+b^i+c^i+d^i.$ Consider $P(x)=x^4-a_1x^3+a_2x^2-a_3x+a_4$ such that $a,b,c,d$ are roots of $P.$ Then the Newton tells us: $b_4-a_1b_3+a_2b_2-a_3b_1+4a_4=0$ Which means that it suffices to show that $n|-a_1b_3+a_2b_2-a_3b_1.$ But by the condition itself we have $n|b_1,n|b_2,n|a_1$ and thus $n|b_4+4a_4= a^4 + b^4 + c^4 + d^4 + 4abcd$ so we are done. $\blacksquare$

This post has been edited 1 time. Last edited by SidVicious, Nov 5, 2016, 10:18 PM

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#4 h Nov 12, 2016, 1:41 PM • 3 Y

Y by Amir Hossein, Adventure10, Mango247

Very easy.From Newton's identity,we have

$s_4-s_3e_1+s_2e_2-s_1e_3+4e_4=0$
consider in modulo $n$ ,it's obvious that $LHS\equiv s_4+4e_4\text{(mod n)}$ and $s_4+4e_4\equiv 0\text{(mod n)}$

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#5 h Jul 19, 2019, 1:33 AM • 7 Y

Y by pavel kozlov, Adventure10, Mango247, Mango247, Mango247, orangesyrup, Amirreza.J

$\begin{split} &(a+b+c+d)(-a^2b-a^2c-a^2d-b^2a-b^2c-b^2d-c^2a-c^2b-c^2d-d^2a-d^2b-d^2c\\ &+abc+abd+acd+bcd)+(a^2+b^2+c^2+d^2)(a^2+b^2+c^2+d^2+ab+ac+ad+bc+bd+cd)\\ =&a^4+b^4+c^4+d^4+4abcd \end{split}$
So we have done :laugh:

:laugh:

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#6 h Apr 12, 2025, 8:22 AM

Y by

Another not bash approach is here.

Since $(a+b)^2 \equiv (c+d)^2 \text{(mod n)}$ and $a^2+b^2 \equiv -c^2-d^2 \text{(mod n)}$ $\implies$ $ab \equiv c^2+d^2+cd \text {(mod n)}$ .

Since $(a^2+b^2)^2 \equiv (c^2+d^2)^2 \text{(mod n)}$ $\implies$ $c^4+d^4+2c^2d^2 \equiv a^4+b^4+2b^2a^2 \equiv a^4+b^4+2(c^2+d^2+cd)^2 \text {(mod n)}$ $\implies$ $a^4+b^4+c^4+d^4+4c^3d+4d^3c+4c^2d^2 \equiv a^4+b^4+c^4+d^4+4abcd \text {(mod n)}$ , and we are done.

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