A number theory problem by Pramita Kastha

If a , b a,b are natural numbers such that a 2 b 2 = 23 a^2 - b^2 = 23 , find a + b a+b .


The answer is 23.

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5 solutions

a 2 b 2 = 23 a^2-b^2=23

( a b ) ( a + b ) = 23 \Rightarrow (a-b)(a+b)=23

Because 23 23 is a prime number, therefore 23 = 23 × 1 23=23\times1

( a b ) ( a + b ) = 23 × 1 (a-b)(a+b)=23\times1

In natural numbers, addition of two numbers is always larger than subtraction:

( a b ) = 1 (a-b)=1

( a + b ) = 23 (a+b)=\boxed{23}

You can stop here unless you want to find the value. If a b = 1 a = b + 1 a-b=1 \Rightarrow a=b+1 , a + b = 23 a = 23 b a+b=23 \Rightarrow a=23-b

b + 1 = 23 b \Rightarrow b+1=23-b

2 b = 24 b = 12 a = 11 2b=24 \Rightarrow b=\boxed{12} \Rightarrow a=\boxed{11}

Akshay Yadav
Aug 28, 2015

a 2 b 2 = 23 a^{2}-b^{2} = 23

( a b ) ( a + b ) = 1 × 23 (a-b)(a+b) = 1 \times 23

Clearly,

( a b ) < ( a + b ) (a-b) < (a+b)

Hence,

( a b ) = 1 (a-b) = 1

( a + b ) = 23 (a+b) = 23

Anurag Motalia
Nov 4, 2015

23 can only be divided by 1 after solving equation. as we solve we would get 12 &11

Given that a 2 b 2 = 23 a^{2} -b^{2} = 23 and that Consecutive squares differ by odd integers , we are looking for n n such that 2 n + 1 = 23 2 n+1=23 . Thus, n = 11 n=11 which means that a = 12 a=12 and b = 11 b=11 . Therefore, a + b = 23 a + b = 23 .

Ian McKay
Sep 3, 2015

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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