How many quadruples of positive integers $(a,b,c,d)$ have property $ab + bc + cd + da = 2016$?

Question

How many ordered quadruples $(a, b, c, d)$ of positive integers have the property that $ab + bc + cd + da = 2016$?

Any hints on a solving strategy or how to approach this problem?

Answer 1

@Anurag has already observed that $(a+c)(b+d)=2016=2^53^27$. We split the factors into $2^x3^y7^z$ and $2^{5-x}3^{2-y}7^{1-z} $ where $0\leq x \leq 5,0 \leq y \leq 2,0\leq z \leq 1$, The number of solutions in positive (nonzero) integers to $x+y=n$ is $n-1$ so number of solutions to this case $(x,y,z)$ is $(2^x3^y7^z-1)(2^{5-x}3^{2-y}7^{1-z}-1)=2017 - (2^x3^y7^z+2^{5-x}3^{2-y}7^{1-z})$ Now summing over all $(x,y,z)$ we get the total number of solutions to be $\sum_{x=0}^5 \sum_{y=0}^2 \sum_{z=0}^1 2017 - (2^x3^y7^z+2^{5-x}3^{2-y}7^{1-z})=72612 -8 \sum_{x=0}^5 \sum_{y=0}^2 (2^x3^y+2^{5-x}3^{2-y}) = 72612-104\sum_{x=0}^5 2^x + 2^{5-x} = 72612-13104=59508$. Thus there are 59508 such ordered quadruplets.

Answer 2

Let $S$ the set of quadruples of positive integers $(a,b,c,d)$ such that $$(a + c)(b + d) = 2016.$$ We have $$ S = \left\{(a,b,c,d)\in\mathbb{N}\,\Big\lvert\, a + c \,\vert\, 2016 \text{ and } b + d = 2016/(a + c) \right\} = \bigcup_{d\,\vert\,2016} S_d $$ where $S_d=\left\{ (a,b,c,d)\in\mathbb{N} \,\Big\lvert\, a + c = d \text{ and } b + d = 2016/d \right\}$.

For a given divisor $d$ of $2016$, the solutions of the equation $a + c = d$ are $$(1,d-1), (2, d-2), \dots, (d-1,1).$$ There are $d-1$ solutions. Also, for each solution of the equation $a + c = d$, the equation $b + d = 2016/d$ has $2016/d - 1$ solutions. Therefore, we have $$\vert S_d \vert = (d - 1)(2016/d - 1) = 2017 - d - 2016/d.$$ As the sets $\{S_d\}_{d\,\vert\, 2016}$ are pairwise disjoints, $$ \begin{align*} \vert S \vert &= \sum_{d\,\vert\,2016} \vert S_d \vert \\ &= \sum_{d\,\vert\,2016} (2017 - d - 2016/d) \\ &= 2017\times\sum_{d\,\vert\,2016} 1 - \sum_{d\,\vert\,2016} d - \sum_{d\,\vert\,2016} 2016/d. \end{align*} $$ Note that $d\mapsto 2016/d$ is a bijective function on the set of divisors of $2016$, so $$\sum_{d\,\vert\,2016} d/2016 = \sum_{d\,\vert\,2016} d.$$

The canonical representation of $2016$ is $2^5\times 3^2 \times 7$. The positive divisors of $2016$ are the integers of the form $2^\alpha \times 3^\beta \times 7^\gamma$ with $0\leqslant \alpha\leqslant 5$, $0\leqslant \beta \leqslant 2$ and $0\leqslant \gamma \leqslant 1$, so the number of positive divisors of $2016$ is $$\sum_{d\,\vert\,2016} 1 = 6\times 3\times 2 = 36,$$ and the sum of the positive divisors of $2016$ is $$ \begin{align} \sum_{d\,\vert\,2016} d &= \sum\limits_{\substack{ 0\leqslant\alpha\leqslant 5 \\ 0\leqslant\beta\leqslant 2 \\ 0\leqslant\gamma\leqslant 1}} 2^\alpha\times 3^\beta\times 7^\gamma \\ &= \biggl(\sum_{0\leqslant\alpha\leqslant 6} 2^\alpha\biggr) \biggl(\sum_{0\leqslant\beta\leqslant 2} 3^\beta\biggr) \biggl(\sum_{0\leqslant\gamma\leqslant 1} 7^\gamma\biggr) \\ &= \frac{2^6-1}{2-1}\times\frac{3^3-1}{3 - 1}\times\frac{7^2-1}{7-1} \\ &= 6552. \end{align} $$ Finally, the equation has $\lvert S \lvert = 2017\times 36 - 2\times 6552 = 59508$ solutions.