can YOU help me with this, because nobody else can figure it out.?

Question

ok, so i have a trig challange problem that ive been stuck on for quite awhile. any helpful hints would be nice, or just how to do one of them. here goes...


Which of the following numbers can-not be expressed as the difference of the sqaures of two integers?


(a) 314159265
(b) 314159266
(c) 314159267
(d) 314159268
(e) 314159269



again, ANY help would be very appreciated. Also, please dont bother commenting if you are going to try and tell me to do my own work, trust me ive tried. ive been stuck on this problem for quite awhile and am running out of ideas, therefore i am going to a worldwide community for help. if you dont have something constructive to say, then please don't waste both of our time. again thank you in advance for any help you may be able to offer me, its VERY needed.

Answer 1

We first note that any odd number can be written in the form 2n + 1 and can be expressed as the difference of perfect squares as follows:(n + 1)^2 − n^2 = n^2 + 2n + 1 − n^2 = 2n + 1.

Also, any number divisible by 4 we can write in the form 4n and can be expressed as the difference of perfect squares as follows:(n + 1)^2 − (n − 1)^2 = (n^2 + 2n + 1) − (n^2 − 2n + 1) = 4n.

So test the numbers by dividing by 4. All odd numbers will have a remainder of 1 or 3. All numbers perfectly divisible by 4 will have a remainder of 0. Stated another way, all integers congruent to 0, 1 or 3 (modulo 4) can be expressed as the difference of perfect squares.

Now we just need to show that all integers congruent to 2 (modulo 4) *cannot* be expressed in such a way. Note that 0^2≡ 0, 1^2≡ 1, 2^2≡ 0, 3^2≡ 1 (mod 4). So perfect squares can only be congruent to 0 or 1 (modulo 4). Thus, differences of perfect squares can only be congruent to 0−0 ≡ 0, 0−1 ≡ 3, 1−0 ≡ 1, 1−1 ≡0 (mod 4). So a difference of perfect squares can never be congruent to 2 (modulo 4). Thus, all integers congruent to 2 (modulo 4) cannot be expressed as the difference of perfect squares.

The answer is 314159266 because when you divide by 4 the remainder is 2 and there is no way to represent this as the difference of perfect squares.

Hope that helped! :-)

Answer 2

Nice, Puzzling! I have to confess, I was stumped too.

* * * * *

A quicker way to do this occurred to me:

Factor x^2 - y^2 to (x+y)(x-y). Now note that:

If x and y are both odd or both even, then (x+y) and (x-y) are both even. Therefore (x+y)(x-y) has to be a multiple of 4.

If either x or y, but not both, is even (i.e. one is even, the other is odd), then (x+y) and (x-y) are both odd. Therefore (x+y)(x-y) has to be odd.

Therefore, the difference of two squares must be either odd or a multiple of 4 - it can't be even but not a multiple of 4 (2, 6, 10, etc.). To determine if a number is even but not a multiple of 4, you only have to check the last two digits, so the number ending in 66 is the correct answer.

Ethan: I wouldn't have gotten this without seeing the other solutions, so don't give me Best Answer! I just thought I'd point out the shortcut. :)

Answer 3

I think the answer is (b) 314159266.

A^2-B^2 = (A+B)(A-B)
314159266 = 157079633*2 (157079633 is a prime number)
(A+B) + ((A-B) = 2A = 157079633 + 2 = 157079635

Since A is an integer 2A cannot be an odd number

Answer 4

http://www.mathscomp.ms.unimelb.edu.au/solutions/solutions02/intsols02.pdf

you will find the solution there .Problem No.5 page 4

Answer 5

LOL, puzzling just copied and pasted!