show that p2 will leave a remainder 1 when divided by 8 if p is a positive odd integer
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Question - Show that p² will leave a remainder 1 when divided by 8 if p is a positive odd integer ?
(2a+1) = 4a + 4a + 1
= 4a(a+1) + 1
We all know that if the variable (a) is even then a+1 is will be odd, or the vice versa. The point is a(a+1) must be even, since one of two consecutive integers has to be even. Therefore, we can use the expression a(a+1) as 2b ( a × a ), for some integer b. Now we have:
= 4*2b + 1
= 8b + 1
=1
Question - Show that p² will leave a remainder 1 when divided by 8 if p is a positive odd integer ?
(2a+1) = 4a + 4a + 1
= 4a(a+1) + 1
We all know that if the variable (a) is even then a+1 is will be odd, or the vice versa. The point is a(a+1) must be even, since one of two consecutive integers has to be even. Therefore, we can use the expression a(a+1) as 2b ( a × a ), for some integer b. Now we have:
= 4*2b + 1
= 8b + 1
=1
kvnmurty:
let p = (2a+1) ..... then (2a+1)^2 = 4 a^2 + 4a +1 ...
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