Can You Do This Without Using WolframAlpha?

[@wesleydeng71]

Just a minor issue. 8:52 the last two terms should be positive since z = -720.

[@sriharikb6781]

z=-720, so -xz and -yz must be positive

[@cobokobo2115]

amazing...kzbin.info/www/bejne/ZmebqGZmiqqpq5Y

[@WerewolfLord]

That was my thought also. (Confirmed.) Fortunately it isn't required here.

[@田村博志-z8y]

Actually, we have 512^3 + 675^3 + 720^3 = 229・467・7621 229 = 50th prime number 467 = 91st prime number 7621 = 968th prime number

[@sendai-shimin]

I thought it must be divisable under 100 (´･ω･`)

[@田村博志-z8y]

@@ensiehsafary7633 Not manual calculation but by the programing.

[@crazy4hitman755]

Using this identity is the coolest thing I have seen

[@leif1075]

Aren't we just looking for the prime factorization of the sum of these three numbers??

[@elkincampos3804]

You must prove that 467 is prime

[@richardfarrer5616]

Yes, but that's trivial and can be left as an exercise. Square root of 467 is between 21 and 22 so we only need prime factors up to 21. We can exclude some trivially 2: last digit not even 3: digits not divisible by 3 5: last digit not 0 or 5 11: sum of odd position digit - sum of even position ones = 4 + 6 - 7 = 3 which isn't a multiple of 11. For the rest (7, 13, 17, 19) either do division or use a trick, especially useful if doing it in your head. Note that for a prime p other than 2 or 5, if p divides 10n then p divides n. Also, if p divides m then p divided m-p, or m+2p etc. Therefore you can say that if 7 divides 467 then 7 divided 467 - 7 = 460. And if 7 divides 460 then 7 divides 46. Since 7 clear doesn't divide 46 then it doesn't divide 467 either. For 13 I would start "if 13 divides 467 then 13 divides 467 + 13 = 480 so 13 divides 48" For 17 you get 17 divides 467 is equivalent to 17 divides 45 For 19 it's a little trickier but note 3x19 = 57 so 19 divides 467 is the same as 19 divides 467 - 57 = 410 Since none of those primes divide 467 it must be prime.

[@하루에두판]

@@richardfarrer5616 thank you for your explanation

[@mryip06]

nice skills

[@bait6652]

anyway to get teh 3 prime factors in terms of sums of 2n3m5k? or rather hte other 2. or 2 sum repr of 2n3m5k

[@MyOneFiftiethOfADollar]

First, thanks for problem. Definitely a challenge without the aid of technology. The usage of “notice that” too frequently does raise suspicions. Not likely that a person would discover this independently in my opinion.

[@dneary]

"tricks" like the factorization of x^3+y^3+z^3-3xyz come up frequently in contest math - particularly in questions related to Viète's formulas - so thinking of that in the context of a sum of 3 squares is not a stretch. It is also a common tactic in contest math problems to find (and search for relationships between) prime factorizations. Certainly a challenging problem, but I think discoverable for a good contest math student.

[@leif1075]

WHY IN GKDS NAME would anyonenever "nktice" that 720 squared would equal that or multiply by 2.thats random and I don't see why anyone would do that

[@failsmichael2542]

These tricks are cute, but utterly useless in the general case.

[@cobokobo2115]

amazing.....kzbin.info/www/bejne/ZmebqGZmiqqpq5Y

[@Deathranger999]

It’s almost like every math contest problem is different and might require different tricks. Of course there’s no general way to factorize the sum of 3 cubes - if there was that would trivialize the problem.

[@jefflittle8913]

"These tricks are cute, but utterly useless in the general case." Sorry, what's the general case? Going to bars and picking up members of the opposite sex?

[@와우-m1y]

asnwer=150

[@createyourownfuture5410]

150 isn't prime

[@mafprivate8841]

First, it's not a prime. Second, clearly the unit digit of the expression is 3, which is definitely not divisible by 150.

[@cobokobo2115]

amazing...kzbin.info/www/bejne/ZmebqGZmiqqpq5Y