With respect to 1001, ..A relative unknown method to test divisibility by 7,11, and 13 involves this. To check divisibility by 2 (check the last digit), by 3 (sum all the digits and see if that is a multiple of 3), and by 5 (last digit is 0 or 5) is well known. Also one can check easily for number 11 too, but to check divisibility by 7 and 13 is not that simple. (although there are many methods- none too simple) Following method, IMO is simpler than most others, as I learned when I was young , as long as you can work with numbers less than 3 digits…(For example given a number 24034 you do quickly ( 034-24 you get 10 and you know the number is neither divisible by 7 or 11 or 13 and a number say
Given a number 123214 you quickly do 214-123=91 and thus the number (123214) is divisibly by 7 and 13… basically if the number is “abc,def,ghi,klm" you do abc-def+ghi-klm – fill in with zero’s on the left if you need…. Basically you reduce a large number to a number of 3 digits which is very easy to handle - I know this is rather academic when one has calculators..)
With respect to:
All I could prove was the sum is odd and atleast one of the numbers is prime and then I gave up. I cheated a little bit and looked up goldbach conjectures, it still did not help me.
V – Nice work, and Yes Goldbach conjecture implies that S is not even (or not like number 13 or 15 etc). The problem was one of the problems in past IMTS (where one has about 1 month to solve and help for books/computers/internet is allowed) so one can expect it to take some time .. I will post a solution here, if there would still be some interest.
With respect to 3^x+4^y = 5^z
Some Hints: when x<0 (negative) – it is easy to prove no solutions…
When x=0 : (that is 1+4^y=5^z) As you say, has only 1 solution. (proof, may be a little harder). I can give hint/proof (or Vriksh can too) if there is interest.
When x>0 : One Hint: Both x and z have to be even.
(Vriksh has proven x is even). (FWIW - There would still be some work, after that. Also remember, if you can find a "general" method - for general case not only for 3,4,5 - there is a $100,000 prize!
With respect to Vriksh’s S and P problem, for n=3 (or greater than 3) one can prove there are infinite solutions (even non distinct).. I already gave one method which generate infinite cases where there are two solutions… It is possible to give method to generate infinite cases where there are 3 (or as many as you want) solutions for given P and Q
For example: n=3
(82, 6162, 36849)
Each give S=43093 and P=18619210116
(No, I did not use a computer! Or even a calculator I started with the method posted in previous message, that is (1,3k,2(2k-1)) (2,k,3(2k-1) and tried to find integer solution when the first number is 6 –
This quickly reduces to finding integer solutions of a eq x^2-41y^2=200
This is known as Brahmgupta equation (If you want to google, Pell equation may get more hits : for example seehttp://www-groups.dcs.st-and.ac.uk/~history/HistTopics/Pell.html
The method I used, I found very interesting. It uses Bhaskara II’s chakravala method along with Brahmgupta’s way to solve basic Pell equation. I still get amazed that Brahmgupta did all that work in 6th century (?)... centuries before Euler/Fermat/Pell.
(The equation gives a series (infinite solutions for x and y each giving a new triplet)