System of Equations

Equation (1): $(a+c)(b+d)=(ab+cd)+(ad+bc)=2371+2346=4717=89 \times 53$ Equation (2): $(a+d)(b+c)=(ab+cd)+(ac+bd)=2371+2670=5041=71 \times 71$

From equation (1) we have: $\text{(1a):} (a+c = 89 \wedge b+d=53) \vee \text{(1b):} (a+c=53 \wedge b+d=89)$ $$ From equation (2) we have: $a+d=71 \wedge b+c=71)$

Combining (1a) and (2) gives: $a=18+b; c=71-b; d=53-b$

$ab+cd=(18+b)b+(71-b)(53-b) = 18b +b^2+3763-124b+b^2=2371$ $\Rightarrow b^2-53b+696=(b-24)(b-29)=0 \Rightarrow b=24 \vee b=29$ $(a,b,c,d) = (42,24,47,29) \vee (47,29,42,24)$

Combining (1b) and (2) gives: $a=-18+b; c=71-b; d=89+b$

$ab+cd=(-18+b)b+(71-b)(89-b) = -18b +b^2+6319-160b+b^2=2371$ $\Rightarrow b^2-89b+1974=(b-47)(b-42)=0 \Rightarrow b=42 \vee b=47$ $(a,b,c,d) = (24,42,29,47) \vee (29,47,24,42)$

So we have a maximum value of $a = \fbox{47}$ .

We denote the equations to be (1), (2), (3) in that order. Then, (2)-(1) gives us $(b-c)(d-a)=299=13\times23$ and (3)-(1) gives us $(b-d)(c-a)=-25=-5\times5$ Let $b-c=x, c-d=y, d-a=z$ , then we have $\begin{aligned} xz&=13\times23\\ (x+y)(z+y)&=-5\times5 \end{aligned}$

Notice that $x-z=(x+y)-(z+y)$ , $x-z$ could only be $\pm 10$ . So $(x, z)$ could be $(23, 13), (13, 23), (-13, -23), (-23, -13)$ . We'll do the first case, and the rest is similar.

If $x=23, z=13$ , then $x+y=5, z+y=-5\Rightarrow y=-18$ . We have $d=a+z=a+13, c=d+y=a+13-18=a-5, b=c+x=a-5+23=a+18$ . Plugging back to equation (1) we obtain $a(a+18)+(a-5)(a+13)=2371$ . Solving the quadratic, we have $a_1=-42, a_2=29$ . Since $a$ is positive, $a=29$ . For the other three cases, we get $a=24, 47$ , or $42$ , respectively. Thus the maximum value of $a$ is $47$ .

Adding the first two equations and factorising, we get $$(a+d)(b+c)=5041=71 \times 71$$ Since they are all positive integers and each factor is more than 1 so a+d=b+c=71 Adding the first and third equations and factorising we get $$(a+c)(b+d)=4717=53 \times 89$$ Adding the second and third rquations and factorising we get $$(a+b)(c+d)=5016$$ Now we consider two cases. $$\text{Case 1: a+c=53 & b+d=89}$$ $$a+d=71 \Rightarrow d=71-a$$ $$a+c=53 \Rightarrow c=53-a$$ $$b+d=89 \Rightarrow b=18+a$$ $$(a+b)(c+d)=5016 \Rightarrow (2a+18)(124-2a)=5016$$ We solve this quadratic in a and get a = 24 or 29.

Case 2: $$\text{Case 1: a+c=89 & b+d=53}$$ $$a+d=71 \Rightarrow d=71-a$$ $$a+c=53 \Rightarrow c=89-a$$ $$b+d=89 \Rightarrow b=a-18$$ $$(a+b)(c+d)=5016 \Rightarrow (2a-18)(160-2a)=5016$$ We solve this quadratic in a and get a = 42 or 47. Therefore looking at all 4 solutions, the maximum value of a is 47.

We use the differences of the equations to get nice factorisations which will help make our lives easier. We subtract the equations to get 3 factorisations:

$(a-c)(b-d)=25$ $\\\\\\\\\\\ (1*)$

$(a-d)(c-b)=299$ $\\\\\\\\\\\ (2*)$

$(a-b)(c-d)=324$ $\\\\\\\\\\\ (3*)$

We now will assume that each linear factor is positive, which implies that $a$ is the greatest positive integer out of the four variables. We now observe $(2*)$ and find that $299=299 \times 1 = 23 \times 13$ . We start by assuming that $a-d=299$ and $c-b=1$ . We use these two equations to make a substitution in $(1*)$ for $b$ and $d$ in terms of $a$ and $c$ respectively:

$(a-c)((c-1)-(a-299))=(a-c)(c-a+298)=-(a-c)(a-c-298)=25$

We now make the substitution $m=a-c$ and solve for $m$ :

$\implies m(m-298)=-25 \implies m^2-298m+25=0$

Solving for $m$ we find that the discriminant is not a perfect square as a result $a-c$ will take on a non-integer value but we are given that $a$ and $c$ are both integers so $a-c$ should also assume an integral value. Since this is a contradiction, we move on to our last factorisation which is $(a-d)(c-b)=299=23 \times 13$ . We assume that $a-d=23$ (Note that we always assign the factor containing $a$ the larger prime factor of 299 since we are maximising $a$ ) and $c-b=13$ . We now use these two equations to substitute for $b$ and $d$ in $(1*)$ once again:

$(a-c)((c-13)-(a-23))=(a-c)(c-a+10)=-(a-c)(a-c-10)=25$

Again we make a substitution $m=a-c$ and solve for $m$ :

$\implies m(m-10)=-25 \implies m^2-10m+25 = 0 \implies (m-5)^2 = 0 \implies m = a-c=5$ This clearly works as $a-c$ takes on an integral value. From this information we now know that $a-c=b-d=5,a-d=23,c-b=13,(b-d)+(c-b)=c-d=a-b=18$ . This gives us $b=a-18, c = a-5,d=a-23$ . Plugging this back in to $ab+cd=2371$ results in:

$a(a-18)+(a-5)(a-23)=2371 \implies a^2-23a-1128 = 0 \implies a = -24$ or $47$ . Therefore the maximal value is $\boxed{a=47}$ .

Taking the difference of these equations, $(a-c)(b-d)=25=5\times 5$ , $(a-d)(c-b)=299=13\times 23$ . So $a-c=b-d=5$ , and $a=d+23,c=b+13$ . Substituting $b,c,d$ into $ab+cd=2371$ , we obtain the solution $a=47, b=29, c=42, 24$ . Hence the answer is $47$ .

From the equations given it is clear that a and b or c and d are odd and other two is even,(ab+cd=2371(odd no) and other two equations are even) .Also considering the average we can find out the range of values.If assume a,b,c,d to be equal in each equations separately,we get values between 34 and 37 .From the above we get the range(2 no less than 34 and 2 no greater than 37).Taking a sample value (odd no) less than 34 and another odd no greater than 37 and multiply both, subtract it from 2371 and factorize the result ,write it as a multiple of two even numbers.Take the four no's as a,b,c,d and substitute the equations .If one of the equation fails,take another combination of odd numbers as stated above and repeat the same procedure until we get all equations satisfied. Then take the highest number as maximum value of 'a'. here,33 43=1419,2371-1419=952=2 2 2 7 17=28 34, 33 34+43 28=2326(not equal to 2670 or 2346 ) take another( 33 45 is required because the remaining cannot be written as a multiple of two even no's ).Do not go for very high or very low values because our equations did not vary widely.After some samples we go through 29 47=1363,2371-1363=1008=2 2 2 2 3 3 7=42 24, considering the other equations we get,47 42+29 24=2670, 47 24+29*42=2346,Thus we get the answers as 47,29,42,24 and out of them highest is 47

From equation 1 and 3

$(ab+cd)^2-(ad+bc)^2=2371^2-2346^2$ $a^2-c^2)(b^2-d^2)=25 \times 53 \times 89$

Check $a^2-c^2$ is factor from $25 \times 53 \times 89$ , after we get $a$ and $c$ , substitution the value to equation 1,2, and 3.

$a^2-c^2=53 \times 89$ (not satisfy)

$a^2-c^2=25 \times 89$ (not satisfy)

$a^2-c^2=5 \times 89$ , (satisfy) , with $a=47$