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Let $a,b,c,d\in N$ and in strictly increasing order such that $$b^2-bd-d^2=a^2-ac-c^2$$ Prove that $ab+cd$ is not a prime.

Approach: $c^2-d^2-bd=a^2-b^2-ac$

$\implies c^2-d^2-bd=a^2-b^2-ac+bc-bc$

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

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

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

$\implies (c+d)(c-d)+b(c-d)=(a-b)(a+b-c)$

$\implies (c+d+b)(c-d)=(a-b)(a+b-c)$

How to processed further.?

This question is similar to Prove ${a^2+ac-c^2=b^2+bd-d^2}$ and $a > b > c > d \implies ab + cd$ is not prime But he used factoring lemma and provided link but that link is not working.

Can you show me how to do that?

Is there other method too to solve this problem?

nonuser
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mathophile
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  • I updated the linked answer to include the cited Lemma – Bill Dubuque Jun 01 '21 at 20:03
  • The "factoring Lemma" is better known as Euler's Four Number Theorem. It is one of many well-known properties that are equivalent to the uniqueness of prime factorizations, e.g. see here and its links. – Bill Dubuque Jun 01 '21 at 20:21
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    took me a while to find any with distinct integers; common target $-6061,$ pairs $ (43,70); (51,71); (61,73); (63, 74) $ for six ordered quadruples such as $(43,51, 70,71)$ – Will Jagy Jun 01 '21 at 21:16
  • pathfinder for Olympiad mathematics – mathophile Jun 02 '21 at 09:24
  • Given that $ a < b < c < d$, we have $ d^2 + bd - b^2 = c^2 + ca - a^2$ with $ d > c > b > a $. Hence, it's identical to the linked solution, using the change of variables $ a \rightarrow d, b \rightarrow c, c \rightarrow b, d \rightarrow a$, so I'm voting to close as duplicate. FYI I added a nice short solution. – Calvin Lin Jun 28 '21 at 23:38

1 Answers1

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This works if $a>b>c>d>0$.

Write $p= ab+cd$ and assume $p$ is prime. Then plug in $a = (p-cd)/b$ and we get: $$\boxed{p(ab-cd-bc) = (b^2+c^2)(b^2-bd-d^2)}$$ and if we plug $c= (p-ab)/d$ we get $$\boxed{p(ab-cd-ad) = (a^2+d^2)(b^2-bd-d^2)}$$

  1. If $p\mid b^2-bd-d^2$

    • If $b^2-bd-d^2 \geq 0$ then $ab+cd \leq b^2 -bd -d^2<b^2<ab$ a contradiction.
    • If $b^2-bd-d^2 < 0$ then $ab+cd \leq d^2+bd -b^2 <d^2+bd <ab +cd$ a contradiction.

  2. If $p\nmid b^2-bd-d^2$, then $p\mid a^2+d^2$ and $p\mid b^2+c^2$ so $$p^2\mid (a^2+d^2)(b^2+c^2) = (ab+cd)^2 +(ac-bd)^2$$ and thus $$p^2\mid (ac-bd)^2\implies p\mid ac-bd \implies ab+cd\leq ac-bd< ac<ab$$ A contradiction again. So $p$ can not be prime.

nonuser
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  • $d^2<cd$ and $ db<ab$ @WillJagy – nonuser Jun 02 '21 at 18:59
  • the first sentce of the question reads $0 < a < b < c < d $ because they are natural numbers "in strictly increasing order" – Will Jagy Jun 02 '21 at 19:07
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    Well I thought it is another way around. – nonuser Jun 02 '21 at 19:09
  • Check out my solution in the linked question. There's a strong similarity with what you're doing. In fact, from your first 2 equations, we get $ (ab+cd)(cd - ad) = (a^2 + d^2 - b^2 - c^2 ) (b^2 - bd - d^2 ) = (ac-bd ) (b^2 - bd - d^2)$, which is what I used. (Was trying to see if I could patch your solution.) – Calvin Lin Jun 29 '21 at 00:36