If g.c.d. (a, b)=1, then g.c.d. (a–b,a+b)= .....,Select a proper option (a), (b), (c) or (d) from given options so that the statement becomes correct.
(a) 1 or 2
(b) a or b
(c) a+b or a–b
(d) 4
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Let g.c.d(a - b, a + b) = d
then, a - b = md and a + b = nd where m,n ∈ N
now, adding both equations.
(a - b) + (a + b) = md + nd = (m + n)d
2a = (m + n)d -----(1)
similarly, subtracting (a -b) from (a +b)
so, (a + b) - (a - b) = (md - nd)
2b = (m - n)d --------(2)
now , g.c.d (a, b) = 1
=> 2 × g.c.d (a , b) = 2 × 1 = 2
=> g.c.d (2a , 2b) = 2
=> g.c.d {(m+n)d, (m -n)d} = 2
[ by putting equations (1) and (2), ]
=> d × g.c.d(m +n , m -n) = 2
hence, g.c.d (m +n, m - n) = 2/d
therefore it is clear that d = 1 or 2
now, taking a instead of m and b instead of n
then, g.c.d(a + b, a -b) = 1 or 2 ,
or, g.c.d (a - b, a + b) = 1 or 2
therefore option (a) is correct
then, a - b = md and a + b = nd where m,n ∈ N
now, adding both equations.
(a - b) + (a + b) = md + nd = (m + n)d
2a = (m + n)d -----(1)
similarly, subtracting (a -b) from (a +b)
so, (a + b) - (a - b) = (md - nd)
2b = (m - n)d --------(2)
now , g.c.d (a, b) = 1
=> 2 × g.c.d (a , b) = 2 × 1 = 2
=> g.c.d (2a , 2b) = 2
=> g.c.d {(m+n)d, (m -n)d} = 2
[ by putting equations (1) and (2), ]
=> d × g.c.d(m +n , m -n) = 2
hence, g.c.d (m +n, m - n) = 2/d
therefore it is clear that d = 1 or 2
now, taking a instead of m and b instead of n
then, g.c.d(a + b, a -b) = 1 or 2 ,
or, g.c.d (a - b, a + b) = 1 or 2
therefore option (a) is correct
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