[Guest]

Determine the probability that for a positive integer N, chosen at random between 1 and 9 inclusively, the number 1111*N is expressible as the difference of two positive perfect squares in at least one way.

[Izzy]

So, randomly thinking.. at 4:30 am, forgive me. The first few perfect squares = 1,4,9,16,25. The differences of these are 3,5,7,9, which can be expressed as 2n+1. So 1111 = 2n+1, n=555. So, the 555th difference is 1111? I guess plug them all in, and it works with whole numbers, doesn't with numbers that aren't integers.

1111 = 555

2222 = 1110.5

3333 = 1666

4444 = 2221.5

5555 = 2777

6666 = 3332.5

7777 = 3888

8888 = 4443.5

9999 = 4999

(It's cool how the numbers themselves are patterns like that.)

So, er. 5/9?

[Izzy]

Actually, I think I'm right. I just used an online perfect square producer (to save me the trouble), and it works for 1111. I'd check higher.. but it won't let me go over 1000, I cba to get a calculator, and I *really* need to go to sleep.

*edit* Well, I'm right for my nondecimal answers. I haven't thought of a way to check for the others yet.

*edit* Google calculator ftw. I'm right about the ones I'm right about. Fairly sure about the rest.

Edited December 31, 2009 by Izzy

[Guest]

the number 1111*N expressed as difference of two positive perfect squares, which means

1111*N = a^2-b^2

= (a+b)(a-b)

so a+b = 1111

a-b = N

only odd numbers satisfy this condition so probability is 5/9 = 0.555555555555555..

[superprismatic]

The probability is 7/9.

Since the difference of two squares, say a²-b²,

can be factored as (a-b)*(a+b) we have that a-b must account for

some of the factors of 1111*N (call their product F) and a+b must

account for the rest (call their product G). So, we have to solve

a-b=F
a+b=G

From this, we know that 2*a = F+G. So F and G must have the same

parity in order for the above linear system to be solvable in the

positive integers. Thus, if N has any factors of 2, it must have

an even number of them -- just one (or an odd number) would force

F and G to be of different parities.

We exhibit one way to express 1111*N for all N for which it is

possible:

1111 = 56²-45²

2222 can't be done: only one 2 in its factorization

3333 = 67²-34²

4444 = 112²-90²

5555 = 78²-23²

6666 can't be done: only one 2 in its factorization

7777 = 89²-12²

8888 = 123²-79²

9999 = 168²-135²

[Izzy]

^Nice.

[superprismatic]

The probability is 7/9.

Since the difference of two squares, say a²-b²,

can be factored as (a-b)*(a+b) we have that a-b must account for

some of the factors of 1111*N (call their product F) and a+b must

account for the rest (call their product G). So, we have to solve

a-b=F
a+b=G

From this, we know that 2*a = F+G. So F and G must have the same

parity in order for the above linear system to be solvable in the

positive integers. Thus, if N has any factors of 2, it must have

an even number of them -- just one (or an odd number) would force

F and G to be of different parities.

We exhibit one way to express 1111*N for all N for which it is

possible:

1111 = 56²-45²

2222 can't be done: only one 2 in its factorization

3333 = 67²-34²

4444 = 112²-90²

5555 = 78²-23²

6666 can't be done: only one 2 in its factorization

7777 = 89²-12²

8888 = 123²-79²

9999 = 168²-135²

"Thus, if N has any factors of 2, it must have an even number of them -- just one (or an odd number) would force F and G to be of different parities." is not true. It should have read "Thus, N may not have just one factor of 2 as this would force F and G to be of different parities."