Cyclically 130

The equations tell us that $(x_1-x_2)(1-x_3x_4) \; = \; (x_3-x_4)(1-x_1x_2) \; = \; 0$ There are four cases to consider:

1. If $x_1=x_2 = a$ and $x_3 = x_4 = b$ then $a + ab^2 = b + a^2b = 130$ , and hence $(a-b)(1-ab) = 0$ . If $a=b$ then $x_1=x_2=x_3=x_4=5$ . If $ab=1$ then $a+b=130$ and hence $x_1=x_2 = 65-8\sqrt{66}$ , $x_3=x_4=65+8\sqrt{66}$ .

2. If $x_1 = x_2$ and $x_1x_2=1$ then either $x_1=x_2=1$ , in which case $x_3+x_4=130$ and $x_3x_4 = 129$ , so that $x_3=1$ and $x_4=129$ , or else $x_1=x_2=-1$ , in which case $x_3+x_4 = 130$ and $x_3x_4 = -131$ , so that $x_3=-1$ and $x_4=131$ .

3. If $x_3x_4=1$ and $x_3=x_4$ then either $x_3=x_4=1$ , in which case $x_1+x_2=130$ and $x_1x_2 = 129$ , so that $x_1,x_2 = 1,129$ , or else $x_3=x_4 = -1$ , in which case $x_1+x_2=130$ and $x_1x_2=-131$ , so that $x_1,x_2=-1,131$ . Neither of these cases is possible.

4. If $x_3x_4 = x_1x_2 = 1$ then $x_1+x_2 = x_3+x_4= 130$ , and hence $x_1=x_3 = 65 - 8\sqrt{66}$ , $x_2=x_4 = 65 + 8\sqrt{66}$ , which is not possible.

Thus the sum of the possible values of $x_4$ is $5 + 65 + 8\sqrt{66} + 129 + 131 \; = \; 394.992307$ which is $395$ to the nearest integer.

If $x_i = 0$ , then we must have $x_j+0 = 130$ for $j \neq i$ . But then, we must also have $130^3 = x_i+ \prod_{j \neq i} x_j = 130$ which is impossible. Hence $x_i \neq 0$ for all $i$ . Let $R = x_1x_2x_3x_4$ , the given condition in the problem means $x_i + \frac{R}{x_i} = 130$ $\Rightarrow x_i^2 - 130x_i +R = 0$ since $x_i \neq 0$ . Now for $i < j$ , we have $x_i^2 -130x_i + R = x_j^2-130x_j +R$ $(x_i-x_j)(x_i+x_j-130) = 0$ Thus, if we set $i = 1$ , we get to the conclusion that either $x_j = x_1$ or $x_j = 130-x_1$ for $j > 1$ . Divide into cases :

i) $x_1 = x_2 = x_3 = x_4 = a$ . Then we have the equation $a+a^3 = 130$ $(a-5)(a^2+5a+26)=0$ The quadratic equation doesn't have real roots since the discriminant is negative. $x_4 = 5$ is the only solution for this case.

For the next cases $x_4 = 130-x_1 > x_1$ , and so $x_1 < 65$

ii) $x_1 < x_2 = x_3 = x_4 = 130-a$ , then we have the equation $a+(130-a)^3 = 130-a+a(130-a)^2 = 130$ From the first equality, we have $(a-(130-a))+(130-a)^2((130-a)-a) = 0$ $(2a-130)(1-(130-a)^2) = 0$ $2(a-65)(1+(130-a))(1-(130-a)) = 0$ $2(a-65)(a-131)(a-129) = 0$ We know $a \neq 65$ , so we either have $a = 131 \Rightarrow 130-a = -1 < a$ or $a = 129 \Rightarrow 130-a = 1 < a$ . No solutions for this case

iii) $a = x_1 = x_2 = x_3 < x_4 = 130a$ . By letting $b = 130-a$ , we arrive to the same set of equations as case ii), and we get $b = 129 \Rightarrow a = 1$ which gives us $(x_1,x_2,x_3,x_4) = (1,1,1,129)$ and $b = 131 \Rightarrow a = -1$ which gives us $(x_1,x_2,x_3,x_4) = (-1,-1,-1,131)$ . We can easily check these two solutions are valid and the sum of all possible $x_4$ in this case $129+131 = 260$

iv) $a = x_1 = x_2 < x_3 = x_4 = 130-a$ , then we have the equation $a+a(130-a)^2 = 130-a+a^2(130-a) = 130$ From the first equality, we have $(a-(130-a))+a(130-a)((130-a)-a) = 0$ $(2a-130)(1+a(130-a)) = 0$ $2(a-65)(a^2-130a+1) = 0$ Again we know that $a \neq 65$ , for the quadratic equation, let the roots be $a_1 < a_2$ . By Vieta's theorem, $a_1 + a_2 = 130$ and $a_1a_2 =1$ so both are them are positive (easy to check they are real numbers). Also we have $a_i(130-a_i) = 1$ Using the above's equality implies $a_i+a_i(130-a_i)^2 = a_i + (130-a_i) = 130$ and also $(130-a_i)+a_i^2(130-a_i) = (130-a_i) + a_i = 130$ and so $a_1,a_2$ satisfies the equation. But only $a_1 < 65$ since their sum is $130$ . By quadratic roots formula, $a_1 = 65-\sqrt{65^2-1} = \frac{1}{65+\sqrt{65^2-1}} < 0.5$ and so $x_4 = 130-a_i \in (129.5,130)$ . Since this is the last case and there is no other non-integer solutions in other cases, the sum for this case is $130$

Hence, the grand total is $5+260+130 = 395$ .

Let x1=a : x2=b : x3=c : x4=d

a+bcd=130 ......(1)

b+cda=130 ......(2)

c+adb=130 ......(3)

d+abc=130 ......(4)

from 1 & 2 : a+bcd = b+cda i.e. (a-b)(1-cd) = 0

Case 1: a=b from 3 & 4 : (c-d)(1-ab)=0 i.e. (c-d)(1-a^2)=0

Case 1.1 : (1-a^2)=0 i.e. a= 1, -1

Case 1.1.1 : a=b=1 ; cd=129, c+d=130 ; c+129/c=130 ; c^2-130c+129=0 ; c=1,129 ; a=b=c=1, d=129 ;

Case 1.1.2 : a=b=-1 ; cd=-131, c+d=130 ; c-131/c=130 ; c^2-130c-131=0 ; c=-1, 131 ; a=b=c=-1, d=131 ;

Case 1.2 : c=d ;                  a(1+c^2)=130 ;                  c(1+a^2)=130 ;                  (a-c) - ac(a-c) = 0 ;                  (a-c)(1-ac) = 0 ;

Case 1.2.1 : a=c means a=b=c=d=5

Case 1.2.2 : ac=1 ; also, a(1+c^2)=130 ; so, c^2-130c+1=0 ; i.e. c=d=65+8 sqrt(66) , a=b=65-8 sqrt(66) ;

Case 2: cd=1 ; (3) c-(4) d ; (c-d)(c+d-130)=0

Case 2.1: c=d reduces to case 1.1

Case 2.2: c+d=130 ; ie c+1/c=130 ; ie c^2-130c+1=0 ; so, c=65-8 sqrt(66), d=65+8 sqrt(66) ; a+b=130 ; 65+8 sqrt(66) + ab (65-8 sqrt(66) ) = 130 ; ab=1 ; So, this reduces to Case 1.2.2 ;

So, in total, we have 4 solutions: a=b=c=-1, d=131 a=b=c=1, d=129 a=b=c=d=5 a=b=65-8 sqrt(66), c=d=65+8 sqrt(66)

So, S=131+129+5+65+8*sqrt(66) =394.9923 =395 (round off)

If $x_4=0$ , then $x_1=x_2=x_3=130$ . But $0+130 \times 130 \times 130 \neq 130$ . So none of the $x_i$ can be $0$ .

Now we can safely multiply each equation with its 'lone' $x$ to get: $(1) \, x_1^2+x_2x_3x_4=130x_1$ $(2) \, x_2^2+x_1x_3x_4=130x_2$ $(3) \, x_3^2+x_1x_2x_4=130x_3$ $(4) \, x_4^2+x_1x_2x_3=130x_4$

$(4)-(1), (4)-(2),$ and $(4)-(3)$ give us: $x_1=x_4 \vee x_1=130-x_4$ $x_2=x_4 \vee x_2=130-x_4$ $x_3=x_4 \vee x_3=130-x_4$

There are now three different cases: All $x_i=x$ , three of the $x_i=x$ and the fourth equals $130-x$ , and two of the $x_i=x$ and the other two equal $130-x$ .

The first case, $x_1=x_2=x_3=x_4=x$ , gives us: $x+x^3-130=0$ $(x-5)(x^2+5x+26)=0$ $x=5 \vee x= -\frac{5+\iota}{2} \vee x=-\frac{5-\iota}{2}$ $x=5$ gives us $x_1=x_2=x_3=x_4=5.$

The second case, $x_1=x_2=x_3=130-x_4=x$ , gives us: $130-x+x^3-130=0$ $x^3=x$ $x=-1 \vee x=0 \vee x=1$ $x=-1$ gives us $x_1=x_2=x_3=-1$ and $x_4=131$ .

$x=1$ gives us $x_1=x_2=x_3=1$ and $x_4=129$ .

The third case, $130-x_1=130-x_2=x_3=x_4=x$ , gives us: $130-x+(130-x)x^2-130=0$ $x^3-130x^2+x=0$ $x=0 \vee x^2-130x^2+x=0$ $x=0 \vee x=65-\sqrt{65^2-1} \vee x=65+\sqrt{65^2-1}$ If $x_3=x_4=65-\sqrt{65^2-1}$ , $x_1=x_2=65+\sqrt{65^2-1} > x_4$ , so $x_4=65+\sqrt{65^2-1}$ is the only viable solution in this case.

We have $x_4=5, 129, 131,$ or $65+\sqrt{65^2-1}$ .

$5+129+131+65+\sqrt{65^2-1} \approx 394.99$ . So the answer to the question is $\fbox{395}$ .

assume S = sum of all distinct values x4 can take

For ease of typing let me refer to x1, x2,x3,x4as a,b,c,and d.

Now it is given that:

1. a+bcd = 130

2. b+acd = 130

3. c+abd = 130

4. d+abc = 130

Multiplying 1 by a and 2 by b and subtracting leads to

a^2-b^2 = 130(a-b)

Therefore either a+b=130 or a=b

Taking other combination of equations (other than 1 and 2) and performing similar operations would lead to a symmetric result of

b+c=130 or b=c

c+d=130 or c=d

a+d=130 or a=d

Using a=b, (Using a=b also covers using b=c, c=d etc. by symmetry)

The equations reduce to a+acd=130, c+a^2d=130 and d+a^2c=130

The last two equations again lead to c=d or a^2=1

c=d => a+ac^2=130 and c+ca^2 = 130 which gives either a=c (giving a=b=c=d ) or ac=1 which along with the now simplified a+c=130 gives

a=b=129.992307, c=d=0.0076927 or a=b=0.0076927 and c=d=129.992307

Solution set = (129.992,129.992,0.00769,0.00769)

d being the highest value

a^2=1 => c+d=130 and (cd=129 from a=1 or cd= -131 from a= -1 )

leading to the solution sets (1,1,1,129) and (-1,-1,-1,131)

Not using a=b, we have a+b=130, b+c=130, c+d=130 and a+d=130 which gives a=b=c=d.

Finally the case a=b=c=d reduces the equations to a+a^3=130

The only real root of which is a=5.

Therefore S= 129.992307 +129 +131 + 5 = 394.9923

a+bcd=130 ......(1) b+cda=130 ......(2) c+adb=130 ......(3) d+abc=130 ......(4)

from 1 & 2 : a+bcd = b+cda i.e. (a-b)(1-cd) = 0

Case 1: a=b from 3 & 4 : (c-d)(1-ab)=0 i.e. (c-d)(1-a^2)=0

Case 1.1 : (1-a^2)=0 i.e. a= 1, -1

Case 1.1.1 : a=b=1 cd=129, c+d=130 c+129/c=130 c^2-130c+129=0 c=1,129 a=b=c=1, d=129

Case 1.1.2 : a=b=-1 cd=-131, c+d=130 c-131/c=130 c^2-130c-131=0 c=-1, 131 a=b=c=-1, d=131

Case 1.2 : c=d a(1+c^2)=130 c(1+a^2)=130 (a-c) - ac(a-c) = 0 (a-c)(1-ac) = 0

Case 1.2.1 : a=c means a=b=c=d=5

Case 1.2.2 : ac=1 also, a(1+c^2)=130 so, c^2-130c+1=0 i.e. c=d=65+8 sqrt(66) , a=b=65-8 sqrt(66)

Case 2: cd=1 (3) c-(4) d (c-d)(c+d-130)=0

Case 2.1: c=d reduces to case 1.1

Case 2.2: c+d=130 ie c+1/c=130 ie c^2-130c+1=0 so, c=65-8 sqrt(66), d=65+8 sqrt(66) a+b=130 65+8 sqrt(66) + ab (65-8 sqrt(66) ) = 130 ab=1 So, this reduces to Case 1.2.2

So, in total, we have 4 solutions: a=b=c=-1, d=131 a=b=c=1, d=129 a=b=c=d=5 a=b=65-8 sqrt(66), c=d=65+8 sqrt(66)

So, S=131+129+5+65+8*sqrt(66) =394.9923

Let's consider the products

Multiplying the first equation by and the second equation by , and comparing, you get

Quite clearly, either , or , for any two indices.

Next, we show that you can't have three of the terms being different from each other, i.e. we can't have and at the same time. This is because if , then , and implies that . Clearly, we must have

This reduces the problem to 4 cases:

1. This is a straightforward cubic equation, and we get

2.

This gives two simultaneous equations:

But don't forget that we also require , and that is the biggest of the lot. Substituting this requirement, you'll find that y = 1 and , or y = -1, .

3.

Again, substituting in our requirement, we get .

1. You'll find that this is identical to case 2, except that there won't be any solution with being the largest.

So the final answer that you need is

I am using a,b,c and d for the notations, we need to find sum of all possible values of d. we have a+ bcd=130 ...(1) b + acd=130...(2) c +abd =130...(3) d+ abc =130...(4)

Subtracting (2) from (1), a-b + cd(b-a)=0 (b-a)(cd-1) =0

Similarly, taking difference of any 2 equations from (1),(2), (3) and (4) We will get the following eqns :

(b-a)(cd-1)=0 (c-b)(ad-1)=0 (d-c)(ab-1)=0 (c-a)(bd-1)=0 (d-b)(ac-1)=0 (d-a)(bc-1)=0

One obvious solution is then a=b=c=d. (Case 1) Another will be a=b=c (case 2) Another will be a=b and c=d. (case 3)

Using these 3 cases and the first 4 eqns, possible values of d are 129,131,5 and 130(approx)

It is given that: 1. a+bcd = 130 2. b+acd = 130 3. c+abd = 130 and 4. d+abc = 130

Multiplying 1 by a and 2 by b and subtracting leads to a^2-b^2 = 130(a-b)

Therefore either a+b=130 or a=b

Taking other combination of equations (other than 1 and 2) and performing similar operations would lead to a symmetric result of

b+c=130 or b=c c+d=130 or c=d a+d=130 or a=d

Using a=b, (Using a=b also covers using b=c, c=d etc. by symmetry)

The equations reduce to a+acd=130, c+a^2d=130 and d+a^2c=130

The last two equations again lead to c=d or a^2=1

c=d => a+ac^2=130 and c+ca^2 = 130 which gives either a=c (giving a=b=c=d ) or ac=1 which along with the now simplified a+c=130 gives

a=b=129.992307, c=d=0.0076927 or a=b=0.0076927 and c=d=129.992307

Solution set = (129.992,129.992,0.00769,0.00769)

d being the highest value

a^2=1 => c+d=130 and (cd=129 from a=1 or cd= -131 from a= -1 )

leading to the solution sets (1,1,1,129) and (-1,-1,-1,131)

Not using a=b, we have a+b=130, b+c=130, c+d=130 and a+d=130 which gives a=b=c=d.

Finally the case a=b=c=d reduces the equations to a+a^3=130

The only real root of which is a=5.

Therefore S= 129.992307 +129 +131 + 5 = 394.9923