Question

If a and b are integers and (a^2)- (b^2)=16 Which of the following cannot be a value of a
A. -5
B. -4
C. 0
D. 4
E. 5

Answer 1

what is the largest \(-b^2\) can be?

Answer 2

8?

Answer 3

@BlackLai note that since \(a,b\) are integers we also have \(a+b,a-b\) be integers since the set of integers is closed under addition

Answer 4

no, it cannot be 8 because \(-b^2\) is negative as \(b^2\) is positive

Answer 5

now observe we can factor \(a^2-b^2=(a+b)(a-b)=16\) so we're looking for two integer factors, namely \(a+b,a-b\), that yield the product \(16\)

Answer 6

consider the possible factorizations of 16:$$1\times16=16\\2\times8=16\\4\times4=16\\-1\times-16=16\\-2\times-6=16\\-4\times-4=16$$

Answer 7

@oldrin.bataku i was thinking only that \(-b^2\leq 0\) and so \(a\) cannot be \(0\) because
\[0-b^2\leq 0\] integers or no integers

Answer 8

not really sure what this problem has to do with integers in any case
if \(x^2-b^2=16\) the surely \(a^2\geq 16\) right?

Answer 9

oops i meant if \(a^2-b^2=16\) etc

Answer 10

since it doesn't matter which we take to be \(a+b,a-b\) we can just go through our list and compute... that being said, however, observe we can find \(a\) given these two via \(a=((a+b)+(a-b))/2\)... in our list of factorizations we then compute possible \(a\):$$1\times16=16\implies a=17/2\\2\times8=16\implies a=5\\4\times4=16\implies a=4\\-1\times-16=16\implies a=-17/2\\-2\times-8=16\implies a=-5\\-4\times-4=16\implies a=-4$$

Answer 11

ignoring the \(\pm17/2\) as those results are not integers we find that \(a\) can be \(\pm4,5\) hence the only option there not possible is \(0\)

Answer 12

@satellite73 that is a much faster approach... clever

Answer 13

thanks

Answer 14

I see

Answer 15

so what answer we end up with ?

Answer 16

would*

Answer 17

C is the only value not possible for \(a\)

Answer 18

i get it now

Answer 19

thank you so much