Number of solutions for a diophantine equation

Question

Given natural integers a,b,c,d such that: $ab=cd$ and $ a+b+c+d=2019.$

How many solutions exist?

From trial and error with smaller numbers I believe $(a+b)$ and $(c+d)$ divide $2019.$ but I can’t prove it or use it to solve the problem. Any help?

I'm not sure how natural a number $0$ is, given how long it took for people to discover it. I've seen it included and not included frequently, and as long as people state their definitions, it's not usually a point of contention. For this particular problem, if we allow $0$, then the only additional solutions we get have $ab=cd=0$, so for each of the $2020$ ways to write 2019 as an ordered sum of $2$ natural numbers, we get 4 solutions, except for $(0,2019)$ and $(2019,0)$, which both yield only two solutions each. So correcting the difference is easy enough. — Aaron, Dec 19 '21 at 17:29
It’s not really a question for debate. What ISO says is irrelevant, too. Either the OP is working in a context where $0$ is a natural number, or OP is working in a context where $0$ is not a natural number. @EricTowers — Thomas Andrews, Dec 19 '21 at 17:37
@ThomasAndrews : I'm not debating. There is the standard usage of "$\Bbb{N}$ and nonstandard usage. It is unsurprising that pre-standardization usage is nonconformant. Post-standardization nonconformant usage is wrong and should be corrected. — Eric Towers, Dec 19 '21 at 17:55
The same methods in the linked dupe apply here (indeed the method in your accepted answers is already explicitly used there). — Bill Dubuque, Jan 09 '22 at 13:50

nonuser · Accepted Answer · 2021-12-19T17:01:58.490

0

$0$ is not a natural number.

We can write $a=xy$ so that $x\mid c$ and $y\mid d$. Then $c=mx$ and $d=ny$ for some integers $m,n$.

So we have $b =mn$ and thus $mn+mx+ny+xy = 2019$ so $$(m+y)(n+x) = 2019$$

Since $2019 = 3\cdot 673$ we have $m+y =3$ and $n+x= 673$ (or vice versa). So we have $$2\cdot 2\cdot 672$$ solutions.

edited Dec 19 '21 at 17:01

answered Dec 19 '21 at 16:55

nonuser

90,026

Perhaps you accepted this answer to soon? You can upvote it though. – nonuser Dec 19 '21 at 16:58
In particular, this answer assume $0$ is not a natural number. – Thomas Andrews Dec 19 '21 at 17:00
If $0$ is a natural number in the context of this question, then total is $$2((3+1)\cdot (673+1) +(1+1)\cdot (2019+1)).$$ – Thomas Andrews Dec 19 '21 at 17:05
1

You should justify the claim "we can write...". Lacking such it is magic, not mathematics. Of course it is ok to link to proofs of this (which occur here in many places). – Bill Dubuque Dec 19 '21 at 17:24
OK @BillDubuque – nonuser Dec 19 '21 at 17:25

score 0 · Answer 2 · answered Dec 20 '21 at 14:09

All valid permutation of a solution $(a,b,c,d)$ will be taken as a single solution. Among $a,b,c,d$ three of them should have the same parity and the other distinct. The case one even and three odd is discarded by the condition $ab=cd$ so we have to solve the case $$(a,b,c,d)=(x,4y,2y,2x)\text { with x odd }$$ It follows the equation $$3x+6y=2019$$ A particular solutions is $(x,y)=(3,335)$ then we have as general solution $$\begin{cases}x=3+2n\\y=335-n\end{cases}$$ Consequently we have $$(a,b,c,d)=(3+2n,1340-4n,670-2n,6+4n)$$
Of the allowed values of $n$ there are $334$ positive, one null and one negative, which gives $\boxed{336}$ possible solutions (with valid permutations excluded; for example for the largest value of $x=671$ we take $(671,4,1342,2)$ and $(671,4,2,1342)$ as the “same” solution)

Number of solutions for a diophantine equation

2 Answers2