None of this nursery school stuff - a proper maths problem. 25$ reward

[jason_t]

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I am a born mathematical genius (solved high school problems at age of 7) but haven't kept up with it lately. My intuition though tells me there are no solutions without quite finding the proof in my head.

So I will offer my 2 cents to lay the foundation for somebody else. When you ^2 a number you basically just double up the number of factors it consist of (i.e. 30=2*3*5 while 900=2*2*3*3*5*5). When you add 68 you add 2*2*17. The number you get after adding 68 has to contain a number of each factor dividable by 5. I think the assymetric nature of 68 containing of 1 factor of 1 kind and 2 of the other messes that up.

No proof, so I don't claim the $25.

EDIT: Just for clarification I know that addition does not mean that you add the factors (that is multiplying), so somebody need to look into the effect of adding the assymetric number of 68.

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This type of reasoning doesn't work.

29 + 3 is divisible two even though neither term is and in fact both terms are prime.

[[censored]]

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My brain hurts[/b].

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FYP

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[Arnfinn Madsen]

You are right Jason, flaw in my intuition, so now you led me to the correct answer. There are an infite number of primes. When you ^2 a prime you end up with infinite amount of numbers. These are basically random since the distribution of primes is random. random+68=random. Thus x^2+68 leads to an infinite amount of random numbers. An infinite amount of random numbers will have infinite numbers that corresponds to y^5.

So, the numbers of solutions are infinite. Send me $25.

[jason_t]

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You are right Jason, flaw in my intuition, so now you led me to the correct answer. There are an infite number of primes. When you ^2 a prime you end up with infinite amount of numbers. These are basically random since the distribution of primes is random. random+68=random. Thus x^2+68 leads to an infinite amount of random numbers. An infinite amount of random numbers will have infinite numbers that corresponds to y^5.

So, the numbers of solutions are infinite. Send me $25.

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Sorry, but that also doesn't work, even on an intuitive level, because the set of natural numbers that is of the form y^5 is very sparse in the set of all natural numbers: a random set of natural numbers (in this case, those of the form x^2 + 68) could avoid all natural numbers of the form y^5.

[Arnfinn Madsen]

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You are right Jason, flaw in my intuition, so now you led me to the correct answer. There are an infite number of primes. When you ^2 a prime you end up with infinite amount of numbers. These are basically random since the distribution of primes is random. random+68=random. Thus x^2+68 leads to an infinite amount of random numbers. An infinite amount of random numbers will have infinite numbers that corresponds to y^5.

So, the numbers of solutions are infinite. Send me $25.

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Sorry, but that also doesn't work, even on an intuitive level, because the set of natural numbers that is of the form y^5 is very sparse in the set of all natural numbers: a random set of natural numbers (in this case, those of the form x^2 + 68) could avoid all natural numbers of the form y^5.

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You were right last time, but now I think you are wrong. There is no sparsity of number that corresponds to y^5. There is infinite Ys, so there is infinite Y^5's. Sooner or later the x^2+68 will hit, and since it gets infinite chances it will hit infinite times.

[gumpzilla]

One trick similar to what I think you were trying to do that sometimes works to prove there are no solutions is to use modular arithmetic. Consider the Diophantine equation x^2 + 1 = 3y. We can prove there are no integers x, y that solve this. How? We know that x^2 + 1 must be divisible by 3. But if x is an integer, then it will be congruent to 0, 1, or 2 mod 3. (This means it will have remainder 0, 1, or 2 when you divide it by 3) No matter which one you pick, you can't get x^2 + 1 to be divisible by 3. So there are no integer solutions. In this case, I don't think this will be very helpful, though.

[mslif]

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X^2 + 68 = Y^5

Find all the solutions or prove there are none.

25$ via neteller for the first correct answer.

(So if you have finite solutions you must list and show they solve the equation and also prove no others work. Or if you have an infinite set X of solutions you must show that all solutions from set X are roots, and prove no others work. Or you must prove there are no solutions)

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I have just spent almost an hour on this. I am not giving up! I am too invested now.....

[Arnfinn Madsen]

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One trick similar to what I think you were trying to do that sometimes works to prove there are no solutions is to use modular arithmetic. Consider the Diophantine equation x^2 + 1 = 3y. We can prove there are no integers x, y that solve this. How? We know that x^2 + 1 must be divisible by 3. But if x is an integer, then it will be congruent to 0, 1, or 2 mod 3. (This means it will have remainder 0, 1, or 2 when you divide it by 3) No matter which one you pick, you can't get x^2 + 1 to be divisible by 3. So there are no integer solutions. In this case, I don't think this will be very helpful, though.

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No, it will not help now. What I forgot, which Jason pointed out, is that prime distribution is random. Thus there does not need to be any logical connection between x and y. Left side of equation is an unsystematic amount of infinite numbers as is right side of equation. Thus they match eachother i.e. infinite/1,000,000,000 times. Since infinite/1,000,000,000=infinite; there is infinite solutions.

[gumpzilla]

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No, it will not help now. What I forgot, which Jason pointed out, is that prime distribution is random. Thus there does not need to be any logical connection between x and y. Left side of equation is an unsystematic amount of infinite numbers as is right side of equation. Thus they match eachother i.e. infinite/1,000,000,000 times. Since infinite/1,000,000,000=infinite; there is infinite solutions.

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Huh? This doesn't make sense and it doesn't reference the given Diophantine equation, so I can only assume you're talking about all equations of this form. As I pointed out, x^2 + 1 = 3y has no solutions with x,y integer. I invite you to produce a counterexample if you think I am wrong.

[Arnfinn Madsen]

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No, it will not help now. What I forgot, which Jason pointed out, is that prime distribution is random. Thus there does not need to be any logical connection between x and y. Left side of equation is an unsystematic amount of infinite numbers as is right side of equation. Thus they match eachother i.e. infinite/1,000,000,000 times. Since infinite/1,000,000,000=infinite; there is infinite solutions.

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Huh? This doesn't make sense and it doesn't reference the given Diophantine equation, so I can only assume you're talking about all equations of this form. As I pointed out, x^2 + 1 = 3y has no solutions with x,y integer. I invite you to produce a counterexample if you think I am wrong.

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You are right in your example. I guess I did not formulate well enough so I will try again:

Those x's that are interesting to look at are prime numbers. All others are fairly uninteresting. When you add 68 to x^2 you completely mess up the factors. Just try i.e. x=5. x^2=25=5*5 while x^2+68=3*31. For different x's the adding of 68 will work differently since this game is about trying to get the left side to be a number consisting of a dividable number of 5 of each factor. In your example the +1 destroys the possibility of reaching a number dividable by 3, here adding 68 is basically throwing all the factors up into the air. A few times they will land right.