14
$\begingroup$

I find nice that $$ 1+5=2 \cdot 3 \qquad 1 \cdot 5=2 + 3 .$$

Do you know if there are other integer solutions to $$ a+b=c \cdot d \quad \text{ and } \quad a \cdot b=c+d$$ besides the trivial solutions $a=b=c=d=0$ and $a=b=c=d=2$?

$\endgroup$
  • 6
    $\begingroup$ One has the family $(a, b, c, d) = (-t + 1, -1, t, -1)$ parameterized by $t \in \Bbb Z$. Of course, given a solution $(a, b, c, d)$, $(b, a, c, d)$ and $(c, d, a, b)$ are also solutions, and these generate other symmetries, too. $\endgroup$ – Travis Willse Jan 5 '16 at 1:33
  • $\begingroup$ Are there any other solutions in positive integers? We need $c^2 d^2-4(c+d)$ to be the square of a positive integer. $\endgroup$ – DanielWainfleet Jan 5 '16 at 2:02
  • $\begingroup$ @user254665 A now-deleted answer essentially resolved the question for the case $a, b, c, d \geq 0$: One could conclude quickly from it that the only such solutions are the ones OP mentions and the appropriate permutations. $\endgroup$ – Travis Willse Jan 5 '16 at 2:05
6
$\begingroup$

First, note that if $(a, b, c, d)$ is a solution, so are $(a, b, d, c)$, $(c, d, a, b)$ and the five other reorderings these permutations generate.

We can quickly dispense with the case that all of $a, b, c, d$ are positive using an argument of @dREaM: If none of the numbers is $1$, we have $ab \geq a + b = cd \geq c + d = ab$, so $ab = a + b$ and $cd = c + d$, and we may as well assume $a \geq b$ and $c \geq d$. In particular, since $a, b, c, d > 1$, we have $a b \geq 2a \geq a + b = ab$, so $a = b = 2$ and likewise $c = d = 2$, giving the solution $$(2, 2, 2, 2).$$ On the other hand, if at least one number is $1$, say, $a$, we have $b = c + d$ and $1 + b = cd$, so $1 + c + d = cd$, and we may as well assume $c \leq d$. Rearranging gives $(c - 1)(d - 1) = 2$, so the only solution is $c = 2, d = 3$, giving the solution $$(1, 5, 2, 3).$$

Now suppose that at least one of $a, b, c, d$, say, $a$ is $0$. Then, we have $0 = c + d$ and $b = cd$, so $c = -d$ and $b = -d^2$. This gives the solutions $$A_s := (0, -s^2, -s, s), \qquad s \in \Bbb Z .$$

We are left with the case for which at least one of $a, b, c, d$, say, $a$, is negative, and none is $0$. Suppose first that none of the variables is $-1$. If $b < 0$, we must have $cd = a + b < 0$, and so we may assume $c > 0 > d$. On the other hand, $c + d = ab > 0$, and so (using a variation of the argument for the positive case) we have $$ab = (-a)(-b) \geq (-a) + (-b) = -(a + b) = -cd \geq c > c + d = ab,$$ which is absurd. If $b > 0$, we have $c + d = ab < 0$, so at least one of $c, d$, say, $c$ is negative. Moreover, we have $cd = a + b$, so $d$ and $a + b$ have opposite signs. If $d < 0$, then since $c, d < 0$, we are, by exploiting the appropriate permutation, in the above case in which $a, b < 0$, so we may assume that $d > 0$, and hence that $a + b < 0$. Now, $$ab \leq a + b = cd < c + d = ab,$$ which again is absurd, so there no solutions in this case. This leaves only the case in which at least one of $a, b, c, d$ is $-1$, say, $a$. Then, we have $-b = c + d$ and $-1 + b = cd$, so $-1 + (- c - d) = cd$. Rearranging gives $(c + 1)(d + 1) = 0$, so we may assume $c = -1$ giving (up to permtuation) the $1$-parameter family of solutions $$B_t := (-1, t, -1, 1 - t), \qquad t \in \Bbb Z,$$ I mentioned in my comment (this includes two solutions, $B_0$ and $B_1$, which are equivalent by a permutation, that include a zero entry). This exhausts all of the possibilities; in summary:

Any integer solution to the system $$\left\{\begin{array}{rcl}a + b \!\!\!\!& = & \!\!\!\! cd \\ ab \!\!\!\! & = & \!\!\!\! c + d \end{array}\right.$$ is equal (up to the admissible permutations mentioned at the beginning of this answer) to exactly one of

  • $(1, 5, 2, 3)$
  • $(2, 2, 2, 2)$
  • $A_s := (0, -s^2, -s, s)$, $s \geq 0$, and
  • $B_t := (-1, t, -1, 1 - t)$, $t \geq 2$.

The restrictions on the parameters $s, t$ are consequences of the redundancy in the solutions we found: $A_{-s}$ is an admissible permutation of $A_s$, $B_{1 - t}$ an admissible permutation of $B_t$, and $B_1$ one of $A_1$.

| cite | improve this answer | |
$\endgroup$
8
$\begingroup$

Here's a solution when $a$, $b$, $c$, and $d$ are positive; I'm not sure how to tackle the more general problem though (although something tells me it's probably similar).

Note that substituting $b=cd-a$ into the second equation yields $$a(cd-a)=c+d\implies a^2-cad+c+d=0.$$ By the Quadratic Formula on $a$, we have $$a=\dfrac{cd\pm\sqrt{(cd)^2-4(c+d)}}2.$$ This can only be an integer when $(cd)^2-4(c+d)$ is a perfect square.

Note that since $(cd)^2-4(c+d)\equiv (cd)^2\pmod 2$, we have $$(cd)^2-4(c+d)\leq(cd-2)^2.$$ This rearranges to $$-4(c+d)\leq -4cd+4\implies 2\geq (c-1)(d-1).$$ This is great news, because it means we actually don't have many cases to check!

CASE 1: $c=1$ (or $d=1$, for that matter).

Plugging back in, we see that for some integer $K$ we have $$d^2-4(d+1)=(d-2)^2-8=K^2.$$ The only case that works here is $d=5$ (since $d=6$ fails and past that point perfect squares differ by at least $9$). By symmetry, $c=5$, $d=1$ also works.

CASE 2: $c=2$ and $d\leq 3$.

Plugging back in to the original expression, we need $$(2d)^2-4(2+d)=(2d-1)^2-9$$ to be a perfect square. Trial and error works since $d$ is small, but here $d=2$ and $d=3$ both work.

CASE 3: $c=3$ and $d=2$.

Oh look, we've already shown this works! (We don't need to check $d=1$ because that was already covered in Case 1.)

Hence the only possible pairs for $(c,d)$ are $(2,3)$, $(2,2)$, $(5,1)$, arrangements. By a similar argument, we see that the same exact possible pairs are the only possible ones for $(a,b)$ as well.

From here, it's not hard to see that $(a,b,c,d)=\boxed{(2,2,2,2),(2,3,5,1)}$ and permutations are the only possible solutions for $a,b,c,d>0$.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ The reason this solution fails for $a,b,c,d<0$ is because then we can't conclude $(cd)^2-4(c+d)\leq(cd-2)^2$. However, it seems likely that under $c+d\leq 0$ we can say $(cd)^2-4(c+d)\geq(cd+2)^2$ and go from there. $\endgroup$ – David Altizio Jan 5 '16 at 2:15
  • $\begingroup$ Could you explain a bit more why $(cd)^2 - 4(c+d) \leq (cd-2)^2$? $\endgroup$ – A.P. Jan 5 '16 at 2:33
  • 1
    $\begingroup$ We know that $(cd)^2-4(c+d)$ has to be a perfect square; the next smallest perfect square is $(cd-1)^2$, so it has to be at most that. But it can't actually be $(cd-1)^2$ for parity reasons - either the LHS is even and the RHS is odd or vice versa. Hence the largest perfect square it could possibly be is $(cd-2)^2$. $\endgroup$ – David Altizio Jan 5 '16 at 3:07
  • $\begingroup$ Again, this assumes $c+d>0$ - in the more general case, we have to probably work with different bounds. $\endgroup$ – David Altizio Jan 5 '16 at 3:09
0
$\begingroup$

I found that, through solving the equations, we will have following (if a>b):

$$a=\frac{cd+\sqrt{c^2d^2-4(c+d)}}{2}$$ and $$b=\frac{cd-\sqrt{c^2d^2-4(c+d)}}{2}$$ I think it will be useful for you to find all of the solutions. Only you should find the solutions for: $$c^2d^2-4(c+d)=m^2$$ where $m$ is any positive integer.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.