Question

How many pairs of natural numbers are there so that difference of their squares is 60

Answer 1

i got only 2 pairs of solutions work, a = 8, b = 2, and a = 16, b = 14

Answer 2

going to post a pick of my solution in a sec.

Answer 3

good morning joemath up all night?

Answer 4

Alright, this solution revolves around that fact that:
\[a^{2} - b^{2} = (a+b)(a-b)\]
Because we are dealing with natural numbers, a+b and a-b are also natural numbers. So we have 2 natural numbers that multiply to 60, we means they must be factors of 60.

Answer 5

morning satelitte :) i dont know if you can call it up all night, i work up around 3 am ish lol

Answer 6

thanks

Answer 7

yikes.
this problem is asking how many pythgorean triples there are with one number = 60

Answer 8

just thought i would mention it

Answer 9

So basically what i did was check the cases. say if a + b = 10, then a-b would have to be 6, and that leads to a valid solution. infact, the only valid solutions are a+b = 10, a-b = 6, and a+b = 30, a- b = 2, so there are 2 pairs of natural numbers that solve that equation.

Answer 10

how did you scan so kwikly

Answer 11

funny story: i bought a new scanner so I could beat satellite to solutions :P (true story!)

Answer 12

cool