Well?

>>8934998
diphantone equation thing so really fucking big
natural numbers
qed.
like really fucking big numbers

>>8934998
https://mathoverflow.net/questions/227713/estimating-the-size-of-solutions-of-a-diophantine-equation

Then multiply the whole thing by 4.

>>8935034
>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

>>8934998
>3 variables
>1 equation
Infinitely many solutions

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>>8935090
>find natural numbers a, b, c such that a + b + c = 3
>3 variables, 1 equation, therefore infinitely many solutions

>>8935092
>>>/highschool/

>>8935095
>getting this mad about being proven wrong

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>>8935092
>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.

>>8935115
ramen nougat? what?

>>8935115
Can i see your worksheet?

>>8935090
>Infinitely many solutions
not if you restrict yourself to natural numbers, which the question does

>>8935092
Thats not whats happening here since the quotients dont need to be natural numbers.

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>>8935092
To the short bus with you

>>8935022
I thought diophantine equtions were like
ax+by=c

>>8934998
Its a=15.something
b=1 c=1

>>8935491
Whole values only

>>8934998
Alon amit showed how to solve a similar equation on quota https://www.quora.com/How-do-you-find-the-integer-solutions-to-frac-x-y+z-+-frac-y-z+x-+-frac-z-x+y-4?encoded_access_token=6bbe8da22f3848c0a530c01b1122540c

>>8934998
wait, is a=b=c, or not?

>>8935594
If they were LHS gives 3/2

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>>8934998
Lol

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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.

>>8935801
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.

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Well that's a plot

>>8935468
Diophantine equations are just equations where you only care about integer solutions.

>>8935844
crazy how math do that except unironically

>>8935670
>positive hole values

>>8935844
what program is this?

>>8934998
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?

>>8936233
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

>>8935355
Is this image implying that tomato is not a fruit?

>>8935844
What software is this? I fucking need it

>>8934998
Can you just take the derivative?

>>8936186
>>8936874
Grapher, a mac program

>>8936291
rather that opposing fruits and vegetables has little to do with biology

>>8935528
posts solution... gets ignored... /sci/ in a nutshell

>>8934998
Simple case of partial fraction decomposition.

>>8935528
Amazing. The best I could do was show that 15 < a/(b+c) < 16

>>8935355
Mixing science with history, chances are you are on mount stupid in more than one topic.

According to
http://ami.ektf.hu/uploads/papers/finalpdf/AMI_43_from29to41.pdf

the smallest solution has 414 digits.

>>8935171
>>8935355
>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

>>8934998
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

>>8937850
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.

>>8937864
>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.

>>8937312
How is integrating going to help you?

>>8935670

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.

>>8935844
Damn that's so nice