show that p square will leave a remainder 1 when divided by 8 , if p is an odd positive integer
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(2a+1) = 4a + 4a + 1
= 4a(a+1) + 1
Since, 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.
=) 4×2b + 1
=) 8b + 1
=) 1
Hope it helps
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(2a+1) = 4a + 4a + 1
= 4a(a+1) + 1
Since, 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.
=) 4×2b + 1
=) 8b + 1
=) 1
Hope it helps
Mark as brainliest
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