If u, v, and w are integers, is u > 0 ?

(1) u = v^2 + 1 --> u = non-negative + 1 = positive. Sufficient.
(2) u = w^4 + 1 --> u = non-negative + 1 = positive. Sufficient.

Answer: D.
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Analyse question stem.
So u, v, w are integers. This means that they cannot be fractions. They can be positive OR negative.
The question is asking whether u is positive.

Statement (1).
u = v^2 + 1
Notice we have an even exponent here. Even exponents can have 2 bases (a positive one and a negative one)
So what values can we put for v: -2, 0, 2 (negative, zero, positive)

In all cases (i.e. for all possible values of v, u will always be positive.)
For example if v=0 ... then 0^2 + 1 = 1 ... 1 > 0
If v = -2 ... then -2^2 + 1 = 5 .... 5 > 0.

SUFF.

Statement (2).
Same as statement 1. we have another even exponent. So negative and positive bases will yield the same result. If we take -2, and 2. -2^4 = 16 and 2^4=16.
In this case u will always be positive.

SUFF.

Answer is D.
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If u, v, and w are integers, is u > 0 ?

(1) u = v^2 + 1
(2) u = w^4 + 1

1) sufficient v^2 is non negative+1(positive) whole equation positive

2) in same sense 2 is sufficient

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