Register
Email: Password:
Forum » Maths Problem

Maths Problem

Narvius 5 years ago
sqrt(a) + sqrt(b) = sqrt(4 + sqrt(7))

Find all real solutions (a, b).

I have no idea, and I somehow doubt that
for all 0 < sqrt(a) < sqrt(4 + sqrt(7)) :: sqrt(b) = sqrt(4 + sqrt(7)) - sqrt(a) therefore for all 0 < a < 4 + sqrt(7) :: b = (sqrt(4 + sqrt(7)) - sqrt(a)) ^ 2
is a sufficient answer.

Halp.
#
Amarth 5 years ago
Sounds right to me?

You can expand the last equation to
b = 4 + sqrt(7) - 2 * sqrt( (4 + sqrt(7)) * a ) + a
but that's about all you can say about it I guess.
#
Narvius 5 years ago
Okay then... it just seems too easy to me.

The more fun one:

Prove that for any array P containing 3^k or more integers, it's possible to pick a subarray S containing (k + 1) items, in a way that for no two (different) subarrays of S the sum of their items are the same.
#
E_net4 5 years ago
Well, you're dealing with subarrays of a subarray there. Have you tried through mathematical induction? Though it's strange that the sum of both subarrays will always be the same if all array items are zero.
#
Amarth 5 years ago
I'm going to guess it's not about arrays but about sets, ie no repetition of integers.

This problem sounds like mathematical induction and/or the pigeonhole principle. Hmm. I don't immediately see it and I don't have much time to think about it right now, but maybe later.
#
E_net4 5 years ago
"Amarth" said:
I'm going to guess it's not about arrays but about sets, ie no repetition of integers.
Now that really changes the whole thing. Let's assume that.

For k = 0, we have a subset of k + 1 = 1 elements. Then we can make an empty subset of this one and another one with this single item... But the sum of both could STILL be 0. Narvius, you'll have to be more specific.
#
Forum » Maths Problem

Post Reply


Your email:
Your name: