If a , b are natural numbers such that a 2 − b 2 = 2 3 , find a + b .
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a 2 − b 2 = 2 3
( a − b ) ( a + b ) = 1 × 2 3
Clearly,
( a − b ) < ( a + b )
Hence,
( a − b ) = 1
( a + b ) = 2 3
23 can only be divided by 1 after solving equation. as we solve we would get 12 &11
Given that a 2 − b 2 = 2 3 and that Consecutive squares differ by odd integers , we are looking for n such that 2 n + 1 = 2 3 . Thus, n = 1 1 which means that a = 1 2 and b = 1 1 . Therefore, a + b = 2 3 .
The difference between any two squares n and (n+1) is (2n + 1).
Then, looking at 23 +/- 1, you have A, B (respectively) 23 + 1 = 24 24/2 = 12 A = 12
23 - 1 = 22 22/2 = 11 B = 11
A + B = 12 + 11 = 23
Interestingly, (a^2 - b^2) always = (a + b) as long as a = (b + 1).
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a 2 − b 2 = 2 3
⇒ ( a − b ) ( a + b ) = 2 3
Because 2 3 is a prime number, therefore 2 3 = 2 3 × 1
( a − b ) ( a + b ) = 2 3 × 1
In natural numbers, addition of two numbers is always larger than subtraction:
( a − b ) = 1
( a + b ) = 2 3
You can stop here unless you want to find the value. If a − b = 1 ⇒ a = b + 1 , a + b = 2 3 ⇒ a = 2 3 − b
⇒ b + 1 = 2 3 − b
2 b = 2 4 ⇒ b = 1 2 ⇒ a = 1 1