If a and b are integers and |b| < a, is (a^2 + b^2)^(1/2) =

If a and b are integers and \(|b| < a\), is \(√a^2+b^2 = a+b\) ?

(1) \(b^a<0\)
(2) |ab| is not a square of an integer

\(|b| < a\) implies that \(a > 0\) where either \(b < 0\) or \(0 < b < a\) .

\(√a^2+b^2 = a+b\)
There \(a+b > 0\) where \(b < 0 < a\) or \(b > 0\).

So, we have \(a > 0\) and \(b < a\) (\(b < 0\) or \(b > 0\))

(1) \(b^a<0\)
implies that \(b < 0\) since \(a > 0\) only if 'a' is odd i.e. 3, 5 and so on.
'b' can take values -1, -2 -3, -4 and so on.

Minimum value of \(a + b = 3 - 1 = 2\)
minimum value of \(√a^2+b^2 = √3^2+(-1)^2 = √10\) Hence NO.
Further calculations would reveal that \(√a^2+b^2\) is always bigger than because \(a^2 > a + b\) always.
Hence \(√a^2+b^2 ≠ a + b\) always. Thus, always a NO answer.

SUFFICIENT.

(2) |ab| is not a square of an integer

\(|ab| ≠ integer^2\) since that would make \(a = -b\) which violates \(|b| < a\)

So |ab| = any integer i.e. 1, 2, 3, 5, 6, 7, 8, 10... excepts perfect squares.
\(|ab| ≠ 1\) since that violates \(|b| < a\) [a = 1 and b = -1]

Thus minimum value of \(|ab| = 2\) where a = 2 and b = -1. So \(√a^2+b^2 = √2^2+(-1)^2 = √5\) and \(a + b = 2 - 1 = 1\). Hence NO.
Or \(|ab| = 3\) where a = 3 and b = -1 So \(√a^2+b^2 = √3^2+(-1)^2 = √10\) and \(a + b = 3 - 1 = 2\). Hence NO.

Thus, \(√a^2+b^2 ≠ a+b\) always.
HENCE NO.

SUFFICIENT.

Answer (D).