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A friend of mine asked me this question yesterday and I tried to solve it using case distinctions which turned out to be rather cumbersome, and he then showed me a very elegant proof which I would like to share as I have not found it anywhere on the internet (please share if you find the name of the problem!).

Let $0<a, b, c, d \in \mathbb{N}$, $ab = cd$. Say whether $a+b+c+d$ is prime and prove it.

My approach was the following: Suppose the sum was prime, call it $p:=a+b+c+d$. Then at least one of our numbers must be odd, since the sum of four even numbers is even. Similarly, the sum of two odd numbers is even, so we can exclude the case of two odd and two even numbers.

Suppose we had exactly one even number, wlog let $d$ even. then $cd$ is an even number since multiplication with even yields even, but $ab$ is odd since the product of odd numbers stays odd, contradiction.

Suppose we had exactly one odd number, wlog let $a$ odd. Then we factor $a, b, c, d$ into primes and find $ab = \Pi_{i} p_i^{n_i}=cd$. Since $c, d$ both are even we have at least two factors 2 in the factorization. This is where I get stuck.

So this is my friend's answer:

Let $p:=a + b + c + d$. Multiply with any of the numbers, wlog $a$: Then $pa=a²+ab+ac+ad$. We then use our assumption $ab=cd$ to substitute $pa=a²+cd+ac+ad$ which we can factor into $pa=(a+d)(a+c)$. Since both these factors are larger than $a$, one of them must divide $p$, hence $p$ is not prime.

EDIT: I love this proof and would like to get better at thinking in this manner. My actual question is: Does this problem have a name and do you know any similar riddles/ problems I could work on?

Bill Dubuque
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arridadiyaat
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  • As you formulated the problem, choosing $a=c=0$ yields opposite results when $b=d=1$ and when $b=d=2$. – Paul Tanenbaum Jun 14 '24 at 10:31
  • @mihaild thanks, I answered it myself below. you can do Q&A type posts on stackexchange, I was only asking for the name of that problem actually, not the solution. my bad – arridadiyaat Jun 14 '24 at 11:14
  • @PaulTanenbaum Thanks, I forgot to put $0<a, b, c, d$. However, I think your example of $b=d=2$ does not work either since $4$ is not prime. – arridadiyaat Jun 14 '24 at 11:15
  • The presented proof works only for positive integers , but you allow arbitary positive real numbers $a,b,c,d$ – Peter Jun 14 '24 at 12:28
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    Unfortunately , such tasks are often about having a good idea rather than using a general scheme. I am not sure how long I would have needed to find this (or another) proof. In particular , in number theory , often clever tricks are worth more than memorizing hundreds of methods or theorems. – Peter Jun 14 '24 at 13:05
  • Worth learning how to prove this simple thing: when $ab=cd,$ there are parameters $g,h, \alpha, \gamma $ such that $a = g \alpha,$ $b = h \gamma,$ $c = g \gamma,$ finally $d = h \alpha.$ Proof: take $g = \gcd(a,c)$ then $a = g \alpha, c = g \gamma$ with $\gcd(\alpha, \gamma = 1.$ So far $b \alpha = d \gamma.$ BUT $\gcd(\alpha, \gamma = 1.$ So, $\alpha$ divides $d,$ take $d = h \alpha.$ ....Finally your $a+b+c+d = (g+h)(\alpha + \gamma)$ – Will Jagy Jun 14 '24 at 16:47
  • The lemma I proved, worth remembering: if $ab=cd$ there are parameters, call them $w,x,y,z,$ with $a = wx, b = yz, c = w y, d = xz$ – Will Jagy Jun 14 '24 at 16:51
  • As I remark in the linked dupe, the Lemma mentioned by @WillJagy in the prior comments is known as the Four Number Theorem. It is the basis of a refinement based approach to unique factorization. See also here. – Bill Dubuque Jun 14 '24 at 18:15

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