Determine all real numbers $a, b, c, d$ for which $$ab + c + d = 3$$$$bc + d + a = 5$$$$cd + a + b = 2$$$$da + b + c = 6$$
Problem
Source: 2021 JBMO TST Bosnia and Herzegovina P1
Tags: algebra, system of equations
MathsSolver007
08.10.2022 00:08
nvm messed up
MathsSolver007
08.10.2022 00:22
Add all the four equations to get $ab+a+b+bc+b+c+cd+c+d+da+a+d=16$ or $(a+1)(b+1)+(b+1)(c+1)+(c+1)(d+1)+(d+1)(a+1)=20$ or $(a+c+2)(b+d+2)=20$ .
Now adding the first 2 equations and the last 2 equations we get that
$(b-d)(a+c-2)=0$. Thus either $a+c=2$ or $b=d$.
We make cases:
Case-1: $a+c=2$
We get that $b+d=3$
Now adding equations 1 and 4, we have that $ab+c+d+da+b+c=9$ or $a(b+d)+b+d+2c=9$ or $3a+2c=6$ since $b+d=3$.
We also have that $a+c=2$
Solving yields $a=2,c=0$. Putting in the values in the other equations give $b=0$ and $d=3$.
We see that $(a,b,c,d)=(2,0,0,3)$ indeed satisfies the system of equations.
Case 2:$b=d$
Putting $b=d$ in equations 1 and 4, we have that $ab+c+d=ad+c+d=3$ and $da+b+c=da+d+c=6$ which is inconsistent. Hence this case doesn't admit any solutions.
Hence the original equation has only solution $(2,0,0,3)$
pinkpig
08.10.2022 01:38
MathsSolver007 wrote:
Add all the four equations to get $ab+a+b+bc+b+c+cd+c+d+da+a+d=16$ or $(a+1)(b+1)+(b+1)(c+1)+(c+1)(d+1)+(d+1)(a+1)=20$ or $(a+c+2)(b+d+2)=20$ .
Now adding the first 2 equations and the last 2 equations we get that
$(b-d)(a+c-2)=0$. Thus either $a+c=2$ or $b=d$.
We make cases:
Case-1: $a+c=2$
We get that $b+d=3$
Now adding equations 1 and 4, we have that $ab+c+d+da+b+c=9$ or $a(b+d)+b+d+2c=9$ or $3a+2c=6$ since $b+d=3$.
We also have that $a+c=2$
Solving yields $a=2,c=0$. Putting in the values in the other equations give $b=0$ and $d=3$.
We see that $(a,b,c,d)=(2,0,0,3)$ indeed satisfies the system of equations.
Case 2:$b=d$
Putting $b=d$ in equations 1 and 4, we have that $ab+c+d=ad+c+d=3$ and $da+b+c=da+d+c=6$ which is inconsistent. Hence this case doesn't admit any solutions.
Hence the original equation has only solution $(2,0,0,3)$
reals, not integers
hi_how_are_u
08.10.2022 02:51
4-1 $da+b+c-ab-c-d=3$ $da+b-ab-d=3$ $d(a-1)+b(1-a)=3$ $(d-b)(a-1)=3$ 2-3 $bc+d+a-cd-a-b=3$ $bc+d-cd-b=3$ $b(c-1)+d(1-c)=3$ $(b-d)(c-1)=3$ $(b-d)(c-1)=(d-b)(a-1)$ $c=2-a$ (1)+(2) $b+d=3$ $a+b-ab=2$ $(2-a)(3-b)+a+b=2$ $b(a-1) = 2(a-2)$ $b(a-1) =a-2$ $(a,b,c,d) = (2,0,0,3)$
MathsSolver007
08.10.2022 08:14
pinkpig wrote: MathsSolver007 wrote:
Add all the four equations to get $ab+a+b+bc+b+c+cd+c+d+da+a+d=16$ or $(a+1)(b+1)+(b+1)(c+1)+(c+1)(d+1)+(d+1)(a+1)=20$ or $(a+c+2)(b+d+2)=20$ .
Now adding the first 2 equations and the last 2 equations we get that
$(b-d)(a+c-2)=0$. Thus either $a+c=2$ or $b=d$.
We make cases:
Case-1: $a+c=2$
We get that $b+d=3$
Now adding equations 1 and 4, we have that $ab+c+d+da+b+c=9$ or $a(b+d)+b+d+2c=9$ or $3a+2c=6$ since $b+d=3$.
We also have that $a+c=2$
Solving yields $a=2,c=0$. Putting in the values in the other equations give $b=0$ and $d=3$.
We see that $(a,b,c,d)=(2,0,0,3)$ indeed satisfies the system of equations.
Case 2:$b=d$
Putting $b=d$ in equations 1 and 4, we have that $ab+c+d=ad+c+d=3$ and $da+b+c=da+d+c=6$ which is inconsistent. Hence this case doesn't admit any solutions.
Hence the original equation has only solution $(2,0,0,3)$
reals, not integers I did not assume that they are integers
pinkpig
08.10.2022 16:40
o oops i'm stupid and can't read properly
rafigamath
25.05.2023 11:31
My solution is the same as @MathsSolver007 so im not gonna post it again
TurandotFromDisney
20.02.2024 17:26
Take $k = a + b + c + d$ and $A = a-1, B = b-1, C = c-1, D = d-1$. One immediately has $$AB = 4-k, \quad BC = 6-k, \quad CD = 3-k, \quad DA = 7-k.$$Hence $ABCD = (4-k)(3-k) = (6-k)(7-k)$ which gives $k = 5$. This allows us to write $$B = -\frac1A, \quad C = -A,\quad D = \frac2A \text{ with constraint } k = A+B+C+D+4 = 5.$$Solving this yields the desired answer.