Here is a solution to this one.
a(a + 1) = b(b + 2) <=> a² + a + 1 = (b + 1)².
For positive a, a² < a² + a + 1 < (a + 1)², implying there is a perfect square between consecutive squares; contradiction.
For negative a, a(a + 1) = p(p - 1), where p = -a, so there are no solutions for p - 1 > 0; i.e., a < -1.
It is then easy to check for solutions with a = -1, 0.
These are: (a,b) = (-1,-2), (-1,0), (0,-2), (0,0).
What an amazing coincidence! It must be coincidence because it stops working when you use any power higher than 5.
It's a consequence of the following even bigger coincidence!
a^n + (a + 4b + c)^n + (a + b + 2c)^n + (a + 9b + 4c)^n + (a + 6b + 5c)^n + (a + 10b + 6c)^n = (a + b)^n + (a + c)^n + (a + 6b + 2c)^n + (a + 4b + 4c)^n + (a + 10b + 5c)^n + (a + 9b + 6c)^n,
where a, b, c are any positive integers and n can be 1, 2, 3, 4, or 5.
I found this in a book...
1 + 6 + 7 + 17 + 18 + 23 = 2 + 3 + 11 + 13 + 21 + 22;
1^2 + 6^2 + 7^2 + 17^2 + 18^2 + 23^2 = 2^2 + 3^2 + 11^2 + 13^2 + 21^2 + 22^2;
1^3 + 6^3 + 7^3 + 17^3 + 18^3 + 23^3 = 2^3 + 3^3 + 11^3 + 13^3 + 21^3 + 22^3;
1^4 + 6^4 + 7^4 + 17^4 + 18^4 + 23^4 = 2^4 + 3^4 + 11^4 + 13^4 + 21^4 + 22^4;
1^5 + 6^5 + 7^5 + 17^5 + 18^5 + 23^5 = 2^5 + 3^5 + 11^5 + 13^5 + 21^5 + 22^5.
Here's another approach.
Consider f(n) = 11 x 14^n + 1, modulo 15.
Then f(n) = 11 x (-1)^n + 1 = 12 or 5 (mod 15).
Hence f(n) = 0 (mod 3) or 0 (mod 5), is greater than 5 for all n, and is always composite.
(f(n) = 12 (mod 15) <=> f(n) = 15t + 12, for some integer t <=> f(n) = 3(5t + 4) <=> f(n) = 0 (mod 3), etc.)
Or you could use induction.
f(0) = 12 is divisible by 3,
f(1) = 155 is divisible by 5.
Then f(n+2) - f(n) = 11 × 14^n(14^2 - 1) = 11 × 3 × 5 × 13 × 14^n is divisible by both 3 and 5, and the result follows by induction.
(f(n) divisible by 3 implies f(n+2) divisible by 3; f(n) divisible by 5 implies f(n+2) divisible by 5.)
A simple proof that no integer of the form 4n-1 can be written as the sum of two squares.
If n is even, say n = 2r, (2r)² = 4r² = 0 (mod 4). (Leaves remainder 0 when divided by 4.)
If n is odd, say n = 2r + 1, (2r + 1)² = 4(r² + r) + 1 = 1 (mod 4.)
So the sum of two squares, mod 4, is always equal to 0 = 0 + 0, 1 = 0 + 1 or 1 + 0, or 2 = 1 + 1; never 3. (Of course 4n - 1 = 3 (mod 4.))
In fact, for the square of an odd integer, we can write (2r + 1)² = 4r(r + 1) + 1.
Since one of r and r + 1 must be even, 4r(r + 1) is divisible by 8.
So the square of an odd integer leaves remainder 1 when divided by 8, which is sometimes useful.
Random fact for the day: All primes of the form 4n+1 can be written as a²+b².
None of the primes of the form 4n-1 can be written in this way. (a, b, n ∈Z)
If you can prove/disprove this, you will be very rich.
Fermat proved this at the beginning of the 17th century! In fact, he proved that all primes of the form 4n+1 can be written as the sum of two squares in exactly one way. The proof is not trivial. However, there is a very simple proof that no primes of the form 4n-1 can be written as the sum of two squares.
haha this is meant to be a trick question...
Then you should have specified a circular orbit!
For an elliptical orbit, both the speed and the velocity change. But for a circular orbit, although the speed is constant, the velocity changes, thus greatly increasing your chances of pulling off the trick!
It's not possible to express y explicitly in terms of x using any of the standard elementary functions. However, there is a formal solution using something called the Lambert W function. This function can be evaluated using mathematical packages such as Maple and Mathematica, but no calculator currently has a button for it. See the references below for details.
Another approach would be to use the Newton Raphson method. If a is an approximation to a root of
f(x) = 10^x - x^10 = 0, then a - f(a)/f'(a) will be a better approximation.
In this case, we have f'(x) = ln(10) * 10^x - 10x^9.
For example, if a = 1.4 is an approximate solution, then
1.4 - (10^1.4 - 1.4^10)/(ln(10) * 10^1.4 - 10*1.4^9) ~= 1.3744 is a better approximation.
This converges quite rapidly. The next two convergents are, to 10 decimal places, 1.3713296532 and
Finally, here's a page which will solve, for example, the equation 10^x = x^10, giving several answers in terms of the Lambert W function (aka the ProductLog function), along with the numeric values.
Who could possibly think that comparing the distance your feet are apart to the length your foot travels would be a number that has digits going on forever ...
Indeed. And who could imagine that that ratio is equal to 4 - 4/3 + 4/5 - 4/7 + 4/9 - 4/11 + ... ? Or imagine that it is equal to sqrt(6/1^2 + 6/2^2 + 6/3^2 + ... ) ?
I'm not sure if I got that correct, based on the point of the idea. Perhaps I got it wrong.
The first column should be in ascending order.
I'm annoyed at this because at a glance it seems easy to prove how it works, but on closer inspection it's not at all!
I agree; it looks trivial, but it's not!
Take the first 10 natural numbers -- that's 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10. Pick any five of the numbers, and write them in a column, in increasing order. Then take the five remaining numbers -- the ones you didn't pick -- and write them in a second column, in decreasing order.
In each row, subtract the smaller number from the larger one, giving you a third column. (See the example below.) Now add the numbers in the third column.
1 9 8
4 7 3
5 6 1
8 3 5
10 2 8
What is your total? Everybody post their total...
You're almost there, mathsyperson!
It's better to leave the expression as A^2 = (a+b+c)(-a+b+c)(a-b+c)(a+b-c)/16. If a, b, c are all odd then the numerator is the product of four odd numbers, and hence is itself odd. So A^2 is not an integer, and neither is A. The same argument applies if two of a, b, c are equal to 2 and the remaining side length is odd.
Milos has already shown that if a = b = c = 2, then A = sqrt(3). If a = 2 and b, c are odd, then, to form a triangle, we must have b = c, with b > 1. (Try drawing a triangle with sides 2, 9, 11!) So we have an isosceles triangle with base 2 and height sqrt(b^2 - 1). b^2 - 1 is not a perfect square for b > 1, so the height is irrational, and hence also the area.
So we've proved a stronger result, a triangle with all sides lengths equal to either 2 or an odd number does not have integral area.