Question

Which of the following numbers cannot be

expressed as the difference of two squares of

the intergers?

314159265
314159266
314159267
314159268

Answer 1

The third option.

Differences of squares of integers cannot end with the digit "7".


Squares can end only with digits 0,1,4,5,6,9. (check it out)

Now, no way by deducting the units digit of two squares would we get the unit's digit '7'.
there you go!

Answer 2

i think you misunderstand the question

Answer 3

Not again??? what does it mean?

Answer 4

you mean that if "a" and "b" are two integers, the |a^2-b^2| {...magnitude} can which of the given options. right?

Answer 5

example

63 = 8^2 - 1^2

Answer 6

yeah thats what i did. what's the answer then?

Answer 7

ofcourse it isnt possible to find the actual nos. we can only find out which is and which isn't.

Answer 8

i mean not possible *practically

Answer 9

the 3rd option can be expressed as the difference of two squares of

the intergers

Answer 10

What's the reason then? I 'll bet my bottom rupee 314159267 can not be.

Answer 11

it can be apoorvk

Answer 12

how? state your logic.

Answer 13

157079634^2 - 157079633^2 = 314159267

Answer 14

My big bad again. *poker face* *palm face*.
You are a genius alright!

so how do you solve this then? Whats the funda behind this?

Answer 15

and yeah my logic had a loophole. :/

Answer 16

no im not genius lol

Answer 17

And yeah, I learnt never ever to bet my bottom buck ever. :|

so what's the logic then?

Answer 18

Notice that
\[2 n+1=(n+1)^2-n^2
\]
Then every odd number is the difference of two squares

Answer 19

n = ((n + 1) / 2 ) ^2 - ((n - 1) / 2 ) ^2

Answer 20

314159267 = ((314159267 + 1)/2)^2 - ((314159267 - 1)/2)^2

Answer 21

amazing. but then, how do you sort out option B and D?

Answer 22

I am not sure if I understand modulo, but I 'll read up on it. Thanks once again! :)

Answer 23

All integers congruent to 2 (modulo 4) cannot
be expressed as the dierence of perfect squares.

Only numbers that can be expressed
as the dierence of perfect squares are those congruent to 0, 1 or 3
(modulo 4).
314159265= 3141592 x 100 + 65 ≡ 65 ≡1 (mod 4)
314159266= 3141592 x 100 + 66 ≡ 66 ≡2 (mod 4)
314159267= 3141592 x 100 + 67 ≡ 67 ≡3 (mod 4)
314159268= 3141592 x 100 + 68 ≡68 ≡0 (mod 4)
314159269= 3141592 x 100 + 69 ≡69 ≡1 (mod 4)
So, of the ve numbers given, only 314159266 cannot be expressed as
the dierence of perfect squares whereas the others can be.