Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $a>b>c>d$ be positive integers and suppose that $${a^2+ac-c^2=b^2+bd-d^2}$$

Prove that $ab+cd$ is not prime? I don't know if this problem is true.

I found that this same problem has also been posted on AOPS.

But I can't prove this problem. Can anyone help me?

share|cite|improve this question
Why complex-numbers ? – Nikita Evseev Mar 27 '13 at 6:04
Your version does not match the problem on AOPS which you refer to. – coffeemath Mar 27 '13 at 6:31
That's supposed to be $ac+bd=(b+d+a-c)(b+d-a+c)\implies {a^2-ac+c^2=b^2+bd+d^2}$ – Inceptio Mar 27 '13 at 6:36
@nikita2: Complex numbers can be used. – Inceptio Mar 27 '13 at 6:43
@Math110: I have edited the question. Have a look if I have made some error in that. And make sure of answering your own question if you're done with that. Community appreciates that.:) – Inceptio Mar 27 '13 at 13:59
up vote 4 down vote accepted

Rewrite as:

$$a^2-b^2+ac-bc=bd-bc+c^2-d^2$$ $$(a-b)(a+b+c)=(c-d)(c+d-b)$$

Since $a>b>c>d$, each of $a-b, a+b+c, c-d, c+d-b$ is positive. By factoring lemma, there exists $w, x, y, z \in \mathbb{Z}^+$ s.t.

$$a-b=wx, a+b+c=yz, c-d=wy, c+d-b=xz$$

Solving for $a, b, c, d$, we get:

\begin{align} 5a=3wx+2yz-wy-xz \\ 5b=-2wx+2yz-wy-xz \\ 5c=-wx+yz+2wy+2xz \\ 5d=-wx+yz-3wy+2xz \end{align}


\begin{align} & 25(ab+cd) \\ & =(3wx+2yz-wy-xz)(-2wx+2yz-wy-xz) \\ & +(-wx+yz+2wy+2xz)(-wx+yz-3wy+2xz) \\ & =5(z^2-wz-w^2)(x^2+y^2) \end{align}


Since $b>c$,

$$-2wx+2yz-wy-xz=5b>5c=-wx+yz+2wy+2xz$$ $$yz>wx+3wy+3xz$$

In particular, $yz>wx+3wy+3xz>3xz$ implies $y>3x$ and $yz>wx+3wy+3xz>3wy$ implies $z>3w$.

Thus $$x^2+y^2>x^2+9x^2>5$$ $$z^2-wz-w^2=(z-\frac{w}{2})^2-\frac{5w^2}{4}>(3w-\frac{w}{2})^2-\frac{5w^2}{4}=5w^2 \geq 5$$

If $ab+cd$ is a prime, then $ab+cd \geq 4(3)+2(1)>5$, then $5(ab+cd)=(z^2-wz-w^2)(x^2+y^2)$ implies that $ab+cd$ divides exactly 1 of $z^2-wz-w^2$ and $x^2+y^2$. However, the term not divisible by $ab+cd$ must necessarily divide $5$, and thus be $\leq 5$. Since both $z^2-wz-w^2>5$ and $x^2+y^2>5$, we obtain a contradiction.

Therefore $ab+cd$ is not prime.

share|cite|improve this answer
oh,It's very nice!Thank you very much. – math110 Mar 28 '13 at 4:27
That was good. +1 – Inceptio Mar 28 '13 at 5:59
@Ivan Loh a small typo in the line above "Since b>c" :) – Vincent Tjeng Mar 29 '13 at 23:51
@VincentTjeng Fixed. – Ivan Loh Mar 30 '13 at 8:17

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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