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11 Mar 2020, 12:13
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How many pairs of positive integers (a, b) satisfy the equation the equation a^2 = 135 + b^2?
A. 2
B. 3
C. 4
D. 5
E. 6
Posted from my mobile device
A. 2
B. 3
C. 4
D. 5
E. 6
Posted from my mobile device
11 Mar 2020, 12:26
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Dillesh4096
\(a^2 = 135 + b^2\)
\(=> a^2 - b^2 = 135 \)
\(=> (a+b)(a-b) = 135\)
Since a and b are positive integers, we must have: \(a + b > a - b\)
Thus, breaking 135 as a product of its 2 factors, we have:
#1. a+b = 135, a-b = 1
#2. a+b = 45, a-b = 3
#3. a+b = 27, a-b = 5
#4. a+b = 15, a-b = 9
Thus, there are 4 pairs
(Note: each pair above is a combination of 2 odd numbers, which when solved, would ensure integer values of a and b)
Alternative approach:
Number of factors of 135 (= \(3^3 * 5^1\)) is: (3 + 1)*(1 + 1) = 8
=> Number of ways 135 can be broken as a product of 2 factors = 8/2 = 4
Note: This would not necessarily work if there was an even number in place of 135 - think why
shameekv1989
11 Mar 2020, 12:44
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sujoykrdatta
sujoykrdatta :- Is it because if a number is even then it might have an Even*Odd combination in which case, solving a+b and a-b will not result in an integer for a and b?
For odd numbers, all the factors are odd and thus a-b and a+b will both be odd numbers and when we solve these equations it will result in an integer for a and b.
