Prove ${a^2+ac-c^2=b^2+bd-d^2}$ and $a > b > c > d \implies ab + cd$ is not prime

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?

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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

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}

Thus:

\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}

$$5(ab+cd)=(z^2-wz-w^2)(x^2+y^2)$$

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.

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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