Dino Dio

When we see questions with $a+b+c+d = 0$ or something alike, we always think to express one of the terms as the others. In this case, let $a = -(b+c+d)$ ----------- (1).

Substitute (1) into the second given equation and we get,

$-(b^{2} +bc + cd + bd)(c^{2} +bc + cd + bd)(d^{2} +bc + cd + bd) + 528^{2} = 0$

Then, notice that we can factorise this into: $-(b+d)(b+c)(c+b)(c+d)(c+d)(b+d) = - 528^{2}$

From this, it is clear that $(b+d)^{2}(b+c)^{2}(c+d)^{2} = 528^{2}$ . Thus, $(b+d)(b+c)(c+d) = 528$ or $-528$ .

From (1), notice for $a$ to be the maximum, $b+c+d$ must be the minimum, (this is the same as $|b+c+d|$ being the maximum). Hence, for max $a$ , $(b+d)(b+c)(c+d) = -528$ --------(2).

Furthermore, observe that the sum of the terms in the product of (2) equal $b+d+b+c+c+d = 2(b+c+d) = -2a$ . Remembering that for max. $a$ , the absolute value of $b+c+d$ needs to be maximise, hence absolute value of $2(b+c+d)$ needs to be the maximum, which in our question is simply $528 +1+1 = 530$ . [Note $|2(b+c+d)| = 530$ when unordered values of $(b,c,d) = (-264, -264, 263)$ . I got these values by letting $b+c = - 528$ and $c+d=b+d = -1$ . But this isn't important to solve the question, since all we want to know is the maximum absolute value of (b+c+d)).

Thus, the minimum value of $b+c+d = \frac{-530}{2} = -265$ , and maximum $a = -(b+c+d) = \boxed{265}$ .

Generalisation

When faced with the same question (with different values of the $528^{2} = x^{2}$ ), the max. $a = \frac{x+1+1}{2}$

Let us generalise this problem, replacing $528$ with $2n$ , where $n$ is a positive integer. We shall show that the maximum possible value of $a$ is $n+1$

From the first equation, we have $a=-(b+c+d)$ . Substituting this value of $a$ into the second equation, we have:

$(-b^2-bc-db-cd)(-c^2-db-cd-bc)(-d^2-cd-bc-db)+n^2=0$

Note that we can factorise $-b^2-bc-db-cd=-(b^2+bc+db+cd)=-(b+c)(b+d)$

Thus we can factorise, with analagous reasoning on the latter two terms, the equation into:

$(b+c)(b+d)(c+d)(c+b)(d+b)(d+c)=n^2$

Which can be further simplified as:

$(b+c)^2(c+d)^2(d+b)^2=n^2$

We wish to find the maximum possible value of a, which occurs when $(b+c+d)$ is at its minimum value, or when $(b+c)+(c+d)+(d+c)=2(b+c+d)$ is at its minimum value. This would occur with $(b+c),(c+d),(d+b)$ being all negative, which would then result in $(b+c)+(c+d)+(d+b)=-n$ Assume to the contrary we have some $(b+c),(c+d),(d+b)$ where at least one of them has a positive value, then $(b+c+d)$ will have a lower value simply by making said value negative.

We now show that the minimum value of $b+c+d$ is obtained when $[(b+c),(c+d),(d+b)]=[-2n, -1, -1]$ or any other permutation thereof, giving $b+c+d=-n$ , which works with one possible solution being $(a,b,c,d)=(n+1,-n,-n,n-1)$ by establishing the following claim:

Claim : Across all triplets of integers $x,y,z$ where their product is $m$ , the largest possible value for $x+y+z$ is $m+2$ , which occurs for $(x,y,z)=(m,1,1)$ or permutatuins thereof.

Proof : We know that for all pairs of integers $(x,y)$ with their product $m$ , the largest possible product occurs when $(x,y)=(m,1)$ . By fixing one of the integers in our triplet $(x,y,z)$ , we can see that at least one of the other two integers must have the value $1$ to maximise their sum. With said integer with value of $1$ , the other two must follow to have values $n, 1$ , which completes the proof.

For our case involving $(b+c)(c+d)(d+b)=-n$ and us wanting the minimum value for $(b+c)+(c+d)+(d+c)=2(b+c+d)$ , we simply invert the signs for $(b+c),(c+d),(d+b)$ , giving us $b+c+d=\frac{-2n-2}{2}=-n-1$ , with $a=-(b+c+d)=n+1$

For this specific example of our generalised problem with $n=\frac{528}{2}=264$ , we thus have our answer to be $264+1=\fbox{265}$

Note that at the beginning of this solution it is specified that we had numbers of form $2n$ , i.e. even numbers, to replace the value of $528$ . We are unable to find any solution in integers for $(a,b,c,d)$ if said number was odd, as the values $(b+c),(c+d),(d+b)$ would all be odd and consequently $2(b+c+d)$ would be odd as well, a contradiction.

What an interesting problem. I was expecting this to be a lot uglier than it ended up being, although this official solution is nowhere near how messy my scratch paper became.

We first attempt to make that nasty product simpler. Note that using the condition $a+b+c+d=0$ we can see that

$\begin{aligned}ab-cd&=ab+ac-ac-cd=a(b+c)-c(a+d)\\&=a(b+c)+c(b+c)=(a+c)(b+c).\end{aligned}$

Similarly, $ac-bd=(a+d)(b+d)$ and $ad-bc=(a+b)(d+b)$ . Therefore, the second equation becomes

$(a+b)(a+c)(a+d)(b+c)(b+d)(c+d)=-528^2.$
Noting once again that $a+b=-(c+d)$ , $a+c=-(b+d)$ , and $a+d=-(b+c)$ , we can simplify this product to

$\begin{aligned}-(a+b)^2(a+c)^2(a+d)^2&=-528^2\\\implies (a+b)(a+c)(a+d)&=528.\end{aligned}$

It remains to maximize the sum of the three factors on the left hand side, as $(a+b)+(a+c)+(a+d)=3a+b+c+d=2a$ , and maximizing this sum will in turn maximize $a$ .

LEMMA: Suppose $x$ , $y$ , and $z$ are positive integers such that $xyz=c$ . Then $x+y+z\leq c+2$ .

Proof. Note that since $x\geq 1$ and $y\geq 1$ , we have

$(x-1)(y-1)\geq 0\implies x+y\leq xy+1.$

Therefore, by applying this repeatedly we have

$x+y+z\leq xy+z+1\leq xyz+2=c+2,$

as desired. Equality holds when $(x,y,z)=(c,1,1)$ or permutations. $\blacksquare$

Applying this lemma to our original problem problem, we have

$2a\leq 528+1+1=530\implies a\leq \boxed{265}.$

Using $a=-(b+c+d)$ the second relation can be rewritten to

$-(b+c)^2(b+d)^2(c+d)^2 + 528^2 = 0$

So we have $|(b+c)(b+d)(c+d)|=528$ .

Using $x=b+c$ , $y=b+d$ and $z=c+d$ we get:

$|xyz|=528$ and $2a=-(x+y+z)$ .

Maximum for $2a=530$ when e.g. $(x,y,z)=(-528,-1,-1)$ ,

so $a_{max} = \frac{530}{2} = \fbox{265}$