I think it is A.

p=8x+5=4(2x)+4+1=4(2x+1)+1
remainder when p is divided by 4 is 1.
II does not tell us anything.

it is D

for first stmt has been explained by alescau already.

for the second stmt, pick up values for the two positive integers.
Note that P is an odd integer so we have to choose the integers such that
one is odd and other is even
thus, odd^2 + even^2 = odd

let's take 1 &2 => P = 5, and remainder is 1
for 6 and 9 => P = 117 and here remainder is 1
for 4 and 13 => P = 185 thus, remainder =1

hence, both are sufficient

A....


for stmt 1,5,13,21,... are the numbers
which gives same remainder 1 when divided by 4....

in case of b

3=1^2+2^2

5=2^2+1^2....

more than one value with stmt 2

thats y.....i suggest A...

tendays2go wrote:
let's take 1 &2 => P = 5, and remainder is 1
for 6 and 9 => P = 117 and here remainder is 1
for 4 and 13 => P = 185 thus, remainder =1

let's take 3^2+9^2=90, R=2
INSUFF


As alescau posted p=8x+5=4(2x)+4+1=4(2x+1)+1
remainder when p is divided by 4 is 1.
I agree with A

Please note P is positive odd integer.

So 3^2+9^2=90 is not possible.

Ans D