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
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
Hello, 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
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 $\displaystyle a^2$ and $\displaystyle b^2$ are even, then both $\displaystyle a$ and $\displaystyle b$ are even.
. . Then $\displaystyle a = 2m$ and $\displaystyle b = 2n$ for some integers $\displaystyle m$ and $\displaystyle n.$
Hence: .$\displaystyle a^2-b^2\:=\:(2m)^2 - (2n)^2 \:=\:4m^2 - 4n^2\:=\:4(m^2-n^2)$
. . Therefore: $\displaystyle a^2-b^2$ is a multiple of 4.
(2) If both $\displaystyle a^2$ and $\displaystyle b^2$ are odd, then both $\displaystyle a$ and $\displaystyle b$ are odd.
. . Then $\displaystyle a = 2m+1$ and $\displaystyle b = 2n+1$ for some integers $\displaystyle m$ and $\displaystyle n.$
Hence: .$\displaystyle a^2-b^2\:=\:(2m+1)^2-(2n+1)^2\:=\:4m^2 + 4m + 1 - 4n^2 - 4n - 1 \:=\:4(m^2+m-n^2-n)$
. . Therefore: $\displaystyle a^2-b^2$ 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: $\displaystyle B:\;314159266$
. . [Its last two digits is "66", which is not divisible by 4.]
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
**
Two consecutive squares always differ by an odd number
. . because: .$\displaystyle (n+1)^2 - n^2\:=\:2n+1$ ... an odd number
Given an odd difference, we can find the two squares.
Example: $\displaystyle d = 37$
Subtract 1 and divide by 2: .$\displaystyle \frac{37-1}{2} = 18$
The two squares are: $\displaystyle 18^2$ and $\displaystyle 19^2$
Given answer-choice $\displaystyle A:\;314,159,265$
. . we can confidently state that: .$\displaystyle 314,159,265\;=\;157,079,633^2 - 157,079,632^2$