Math, asked by vivekrathor200316, 4 months ago

Let a, b, c be three distinct positive integers such that the sum of any two of them is a
perfect square and having minimal sum, a + b + c. Find thesum.​

Answers

Answered by amitnrw
2

Given : a,b,c be three distinct positive integers such that the sum of any two of them is a perfect square    and having minimal sum, a + b + c

To Find : a + b  + c

Solution:

a   + b  = k²

a  + c  = m²

b  + c   = n²

=> b - c  = k² - m²

=> b + c  =  n²

=> 2b =  k² - m² + n²

=> b =  (k² - m² + n²)/2

a =      (k² + m² - n²)/2

c =     ( -k² + m² + n²)/2

=>k² , m² , n²  all 3 has to be even   or 2 odd and 1 even

and sum of any two > third  as a , b & c are positive

with some hit and trial  Earliest possible combination is

k² = 25

m² = 36

n² =  49

Hence a , b & c are

6 ,  19  , 30

a + b + c = 6 + 19 + 30 = 45

minimal sum, a + b + c.  = 45

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Answered by Anonymous
0

Given :-

a,b,c be three distinct positive integers such that the sum of any two of them is a perfect square    and having minimal sum, a + b + c

To Find : a + b  + c

Solution:-

a   + b  = k²

a  + c  = m²

b  + c   = n²

=> b - c  = k² - m²

=> b + c  =  n²

=> 2b =  k² - m² + n²

=> b =  (k² - m² + n²)/2

a =      (k² + m² - n²)/2

c =     ( -k² + m² + n²)/2

=>k² , m² , n²  all 3 has to be even   or 2 odd and 1 even

and sum of any two > third  as a , b & c are positive

with some hit and trial  Earliest possible combination is

k² = 25

m² = 36

n² =  49

Hence a , b & c are

6 ,  19  , 30

a + b + c = 6 + 19 + 30 = 45

minimal sum, a + b + c.  = 45

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