A Diophantine Equation With Factorials
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a•6! = b^2
a•(3^2 • 4^2 • 5) = b^2
Let a= 5(x^2)
(3•4•5•x)^2 = b^2
(60x)^2 = b^2
So b=60x, a=5(x^2)
The exponents of the prime factors of a perfect square are even !
6!=1*2*3*4*5*6=6*4*5*6=2*2*6*6*5=12*12*5
So if a*6!=b^2 with natural numbers a and b, a must be dividable by 5, because the lleft side of the equation is diviidable by 5, so b must be diviidable by 5 and b^2 must be dividable by 5^2.
So a=5 and b=60 is a solution, also a=5 and b=--60. The others solutions are a=5*c^2 with a whole number c and b=60*c. There is no solution with a negative whole number,because the right side of the equationis can never be smmaler than zero.
B squared is equal to b x b
So a x 6! = b x b
Idk how to explain my working out but you can see 6! = a is a solution, which would also mean that b is equal to both a and 6!
Therefore 6! which equals 720 is a solution
6 !
= 2 * (3!) * (3! ) * 4 * (5 !) ^2 * 6
= ( 4 * (3)! * (5!)) ^2 * 3
Hereby
a * (6!) = b^2 implies
a * 3 * ( 4 * (3!) * (5!)) ^2 = b^2
Hereby
b = 4 * (3!) * (5!) * 3 = (2!) * (3!) * (6!)
a = 3
Well, actually that is not quite true. a must be a multiple of 5, agreed. But it might be a multiple of any prime number greater than 5 raised to the power of 2n, where n is an integer. Every prime number might appear or not as a multiple and n can be different for every prime number...
a solution is a=720, b=720
Nice!
you are very intense
What does that mean?
I read "Dopamine equation" and wondered what kind of property has this expresion that makes you so happy lol
Equations always make me happy hehehe 😜
Very cool and original ! Thank you !
Actually any a of the form 5^(2m+1)*c² will work.
Same as my solution
Why is c positive?
At first glance: 6! is 2^4 * 3^2 * 5^1. All perfect squares have all prime factors to even exponents. So any number of the form 2^(2k ) * 3^(2m) * 5^(2n-1) * p^2 (where k, m, n and p are non-negative integers) should accomplish this. (Edit: actually, we can simply say 5^(2n-1) * p^2, because the 2 and 3 can be rolled into the p.)
It is much simpler .. every digit in the faculty must be compensated by something to give a square. The 6 by 2*3 to give 6*6, the 4=2x2 is a square already so stripe away 2,3,4 and 6 (2*3*4*6 is square) on the faculty site and they need not be compensated anymore , the 5 cannot be compensated, so it must have factor 5 to give a square 5x5. So a = 5 and b=sqrt(4) *5*6 =60 . I did it without paper ,by head.
If a = (1/20) the b would be ±6 OR if a=(1/5) b= ±12, so I think there will be infinitely many solutions.
a = 5, b = 60
😊😊😊👍👍👍🎉🎉🎉
5a x 12^2 = b^2
a = 5^(2n-1), b=12x5^n, non negative int n
8:14 I can hear Any Math saying "Nice".
a=b=6! Surely?
That’s a good one!
Nice
Thanks
6!=720=5*144...a=5,a=20,a=45...a=80...
i did this in my head in 1 minute as i looked at the thumbnail
😲
yes!!!!! 👌
❤
Very nice!
Thanks!
Why do you ramble on
What do you mean?
@@SyberMath i mean that you don’t really get to the point of your problems
Bravo. It's not difficult, but it's interesting how you parametrize the solutions with the integers