Well?
Well?
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diphantone equation thing so really fucking big
natural numbers
qed.
like really fucking big numbers
mathoverflow.net
Then multiply the whole thing by 4.
>Then multiply the whole thing by 4.
explain how multiplying a, b and c by any common factor will ever change the value of that expression
>3 variables
>1 equation
Infinitely many solutions
>find natural numbers a, b, c such that a + b + c = 3
>3 variables, 1 equation, therefore infinitely many solutions
>>>/highschool/
>getting this mad about being proven wrong
>infinitely many solutions
>you don't know a single set of them
I put it into maple and couldn't get an answer so I'm gonna guess there aren't any human-findable solutions. Unless you're Ramanujan I suppose.
ramen nougat? what?
Can i see your worksheet?
>Infinitely many solutions
not if you restrict yourself to natural numbers, which the question does
Thats not whats happening here since the quotients dont need to be natural numbers.
To the short bus with you
I thought diophantine equtions were like
ax+by=c
Its a=15.something
b=1 c=1
Whole values only
Alon amit showed how to solve a similar equation on quota quora.com
wait, is a=b=c, or not?
If they were LHS gives 3/2
Lol
First thing to do is to realize that multiply a, b, and c by a constant multiple does not change the value of that sum. Next, if you just let a be a rational p/q, and set b=c=1, then solving for a like a quadratic will yield an irrational result. This implies that no two numbers can equal each other. Once you have this, do the substitution in the pic and get to the reduction I did, taking note that if a, b, and c are whole numbers, then so are x, y, and z. Once you get here, this fraction must reduce over x,y, and z in order to yield a while number. To divide by x, all terms with an x are divisible, leaving only yz^2+zy^2. Lastly, this term is divisible by both y and x, and from our co-prime rule and inequality, this term is indivisible by x. therefore the fraction cannot be reduced to a whole number, and therefore 16 is unobtainable using whole numbers. Fuck off with your Jewish tricks.
Nvm, I'm fucking retarded. That whole indivisibility thing is false. If x = 10, then (3)(15)^2+(15)(3)^2 is perfectly divisible by 10 and will yield 81.
Well that's a plot
Diophantine equations are just equations where you only care about integer solutions.
crazy how math do that except unironically
>positive hole values
what program is this?
c needs to be about 16 times larger than a+b
cheat a little and say a=0 is non-negative whole number
if b=1 c=15 we get 1/15 + 15/1 (missing 14/15ths)
a = 1 b = 1 c = 29 gives 2/30 + 29/2
how to pick values of a,b,c that give whole number results?
a+b = 8 b+c = 64 a = 2 b = 6 c = 58
a+c = 60
6/60 + 2/64 + 58/8
I think 16 is too high to get the remainders close to summing to a whole
Is this image implying that tomato is not a fruit?
What software is this? I fucking need it
Can you just take the derivative?
Grapher, a mac program
rather that opposing fruits and vegetables has little to do with biology
posts solution... gets ignored... Veeky Forums in a nutshell
Simple case of partial fraction decomposition.
Amazing. The best I could do was show that 15 < a/(b+c) < 16
Mixing science with history, chances are you are on mount stupid in more than one topic.
According to
ami.ektf.hu
the smallest solution has 414 digits.
>having no reading comprehension
The point is that "3 variables and 1 equation" alone does not necessarily lead to the conclusion that there's infinitely many solutions when you're dealing with natural numbers
I can report that there are no solutions if a, b, and c are each between 1 and 5000.
I determined that by exhaustive search.
The closest I got was:
a=124 b=139 c=4192 result = 16.00000004536067
If youre using a computer try up to one millioo
I have a more interesting result now. In searching a, b, and c each between 1 and 5000, none of the results is an integer.
The closest I got was:
a=37 b=4278 c=4781 result=1.999999999904425
So now I want to broaden the challenge: you are now allowed to put any positive integer on the right side of the equation.
>If youre using a computer try up to one millioo
That will be tough. It took 10 minutes for me to go up to 5000. The running time increases by the cube of the number, so going to 1 million would take about 8 million times longer.
How is integrating going to help you?
Yeah, that doesn't look too promising. There's a lot of different irrational factors all over the place in there. The chance of getting an integer for c seems pretty slim.
Damn that's so nice