Karnataka regional mathematical olympiad questions?

Question

!) Find all primes p such that 4p^2+1 and 6p^2+1 are both primes?
2) Find all real x>0 such that x-[x], [x],x (where [x] denotes the greatest integer not greater than x) form a geometic progression./

Answer 1

Wow, these are hard. I think I can solve the first one however...

First consider integers that are one more than a multiple of 5, i.e. p = 5k + 1. Then 4(5k + 1)^2 + 1 cannot be prime: 4(5k + 1)^2 + 1 = 100k^2 + 40n + 5 = 5(20k^2 + 8k + 1), which is a composite number divisible by 5 for any k > 0. Also, if k = 0, then p = 1, which we'll rule out because 1 is not considered a prime number.

The logic is similar for integers that are one less than a multiple of 5, i.e. p = 5k - 1: it follows that 4(5k - 1)^2 + 1 cannot be prime: 4(5k - 1)^2 + 1 = 100k^2 - 40n + 5 = 5(20k^2 - 8k + 1), which is a composite number divisible by 5 for any k > 0. Also, if k = 0, then p is negative, so there are no solutions in this set either.

Now consider integers that are two more than a multiple of 5, i.e. p = 5k + 2. For this set, we'll prove 6p^2 + 1 cannot be prime: 6(5k + 2)^2 + 1 = 150k^2 + 120n + 25 = 5(30k^2 + 24k + 5), which is a composite number divisible by 5 for any k >= 0.

Using similar logic for integers that are two less than a multiple of 5, i.e. p = 5k - 2: 6(5k - 2)^2 + 1 = 150k^2 - 120n + 25 = 5(30k^2 - 24k + 5), which is a composite number divisible by 5 for any k >= 0.

In conclusion, the only numbers p for which both 4p^2 + 1 and 6p^2 + 1 can both be prime must be divisible by 5. Your problem states that p must be prime, which rules out 10, 15, 20, etc. We only need to test p = 5, yielding 101 and 151, both prime numbers.

So, there is only one number: 5.

* * * * *

OK, on the second problem, I think I see a solution:

Let a be the integral part of x and b the fractional part. Then x = a + b, [x] = a, and x - [x] = b. The geometric series becomes: b, a, a+b, where a must be in integer and 0 < b < 1. Note that the third term must be the sum of the first two terms.

Now, for this to be a geometic series, the second term has to be br and the third term br^2 for some value of r. Therefore, since the third term is the sum of the first two terms, br^2 = br + b => r^2 - r - 1 = 0 => r = (sqrt(5) + 1)/2. (Note this is the Golden Ratio.)

Also, since br must be an integer, b = k/r for some integer k. Since 1/r = (sqrt(5) - 1)/2, it follows that b = k * (sqrt(5) - 1)/2. However, if k >= 2, then b > 1, which violates the way we defined b. Therefore, k = 1 is required. We are ready to sum up what we know:

The first term of the geometric series is b = (sqrt(5) - 1)/2

The second term of the geometric series is a = 1

The third term of the geometric series is (sqrt(5) + 1)/2

Or, going back to the original statements, [x] = 1, x = (sqrt(5) + 1)/2

(Break them up next time... too much work for one problem. :) )

Answer 2

this is a guess
for second part
take x as
0.3, 0.9,0.27,0.81,....
or
0.2,0.4,0.8,0.16,....
their can be many more