# Difference of the squares of two integers

• Oct 18th 2006, 01:36 PM
ceasar_19134
Difference of the squares of two integers
Which one of the following numbers cannot be expressed as the difference of the squares of two integers?
A:314159265
B:314159266
C:314159267
D:314159268
E:314159269

For this one, I'm just completley clueless.
• Oct 18th 2006, 01:57 PM
ThePerfectHacker
Quote:

Originally Posted by ceasar_19134
Which one of the following numbers cannot be expressed as the difference of the squares of two integers?
A:314159265
B:314159266
C:314159267
D:314159268
E:314159269

For this one, I'm just completley clueless.

Numbers of the form 4k,4k+1,4k+3 can always be expressed as a difference of squares.
Remove the odd numbers for they can always be expressed as difference of squares are you have,
314159266
314159268
Divide them by 4 and find the one which leaves a remainder of 2.
Since both are even they leave either no remiander or 2.
Thus, find the number not divisible by 4.
The trick is to look at the last two digits,
66----> Not Divisible
68----> Yes Divisible
Thus, the answer is,
314159266
• Oct 18th 2006, 02:02 PM
CaptainBlack
Quote:

Originally Posted by ceasar_19134
Which one of the following numbers cannot be expressed as the difference of the squares of two integers?
A:314159265
B:314159266
C:314159267
D:314159268
E:314159269

For this one, I'm just completley clueless.

The prime factorisation of 314159266 is:

314159266=2x157079633

Now if N=314159266 is the difference of two square integers there exists
integers m > n such that:

N = m^2-n^2 = (m+n)(m-n).

Now if (m-n) is even then both n and m are even or both are odd. But as
we can see from the factorisation (m-n)=2, and (m+n)= 157079633,
that is one of (m+n) and (m-n) is odd and the other even which is not
possible so N is not expressible as the difference of two squares.

RonL
• Oct 27th 2006, 10:21 AM
ceasar_19134
Quote:

Originally Posted by ThePerfectHacker
Numbers of the form 4k,4k+1,4k+3 can always be expressed as a difference of squares.
Remove the odd numbers for they can always be expressed as difference of squares are you have,
314159266
314159268
Divide them by 4 and find the one which leaves a remainder of 2.
Since both are even they leave either no remiander or 2.
Thus, find the number not divisible by 4.
The trick is to look at the last two digits,
66----> Not Divisible
68----> Yes Divisible
Thus, the answer is,
314159266

If I may ask, what does k stand for?

Edit: I didnt know where the quote button would put it, sorry.
• Oct 27th 2006, 10:46 AM
CaptainBlack
Quote:

Originally Posted by ceasar_19134
If I may ask, what does k stand for?

Edit: I didnt know where the quote button would put it, sorry.

any integer, as k varies you get all numbers which leave remainders
of 0, 1 or 3 from 4k,4k+1,4k+3.

RonL
• Oct 27th 2006, 10:57 AM
ceasar_19134
So we subsitute the numbers in as k and then what?

And why does the remainder have any significance?

My teacher does not bother teaching us how to solve any of these problems.
• Oct 27th 2006, 11:21 AM
Soroban
Hello, ceasar_19134!

Quote:

Which one of the following numbers cannot be expressed
as the difference of the squares of two integers?
A: 314159265
B: 314159266
C: 314159267
D: 314159268
E: 314159269

It can be shown that any odd number can be expressed as the difference of two squares. **

The only suspects are B and D.

To have an even difference of squares,
. . either (1) both squares are even or (2) both squares are odd.

(1) If both $a^2$ and $b^2$ are even, then both $a$ and $b$ are even.
. . Then $a = 2m$ and $b = 2n$ for some integers $m$ and $n.$
Hence: . $a^2-b^2\:=\:(2m)^2 - (2n)^2 \:=\:4m^2 - 4n^2\:=\:4(m^2-n^2)$
. . Therefore: $a^2-b^2$ is a multiple of 4.

(2) If both $a^2$ and $b^2$ are odd, then both $a$ and $b$ are odd.
. . Then $a = 2m+1$ and $b = 2n+1$ for some integers $m$ and $n.$
Hence: . $a^2-b^2\:=\:(2m+1)^2-(2n+1)^2\:=\:4m^2 + 4m + 1 - 4n^2 - 4n - 1 \:=\:4(m^2+m-n^2-n)$
. . Therefore: $a^2-b^2$ is a multiple of 4.

We have just established a theorem.
. . The difference of two squares is either odd or a multiple of 4.

We can now "eyeball" the list and determine the answer.

The even choice not divisible by 4 is: $B:\;314159266$
. . [Its last two digits is "66", which is not divisible by 4.]

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**
Two consecutive squares always differ by an odd number
. . because: . $(n+1)^2 - n^2\:=\:2n+1$ ... an odd number

Given an odd difference, we can find the two squares.

Example: $d = 37$

Subtract 1 and divide by 2: . $\frac{37-1}{2} = 18$

The two squares are: $18^2$ and $19^2$

Given answer-choice $A:\;314,159,265$
. . we can confidently state that: . $314,159,265\;=\;157,079,633^2 - 157,079,632^2$