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/math/ - Mathematics


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25 Dec 2021Mathchan is launched into public

10 / 3 / 9 / ?

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What I said in the subject. It's so easy to prove. Search the problem and solve it, simple as that.
>>
>>512
>please... please search this obscure math equation online so that we could pwn you on Keizeer Soze
>>
>>512
then do it?
>>
>>512
In
n!+1=m2 n!+1=m^2
, m can be rewrite into
m˙+n \dot{m}+n
and
n!+1=(n+m˙)2 n!+1=(n+\dot{m})^2
have the same integer solution of
n!+1=(n+(n+1)2)2n!+1=(n(n1)2)2 n!+1=(n+(n+1)^2)^2 \lor n!+1=(n-(n-1)^2)^2
for
n>1 n>1
>>

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>>512
its already proven get with the times smh ;^)

https://www.researchgate.net/publication/316110458_SOLVING_THE_BROCARD_PROBLEM
>>
n==4 and m==5
Proof
4!=1*2*3*4=2*3*4=6*4=24
24+1=25
sqrt(25)=5
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>> 548
It's obviously not a real proof. It's written in MS Word (probably) by someone with elementary knowledge on number theory. It contains so many mistakes, I don't even want to read it.
>>
>> 555
Lmfao proof to the contrary by
>wrong formatting software
and
>inadequate credentials of the author

at least referring to the mistakes would be worth something, the other two points have zero relevance to how true any of the statements are
>>

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>>512
This is a very safe statement to make when you don't use "iff", I'd be willing to wager it's true and factpilled
>>
>>512
I'm a retarded dumb zoomer and I solved it so.... yeah my iq is like 120+ or sumn
>>
iq autism