Any integers square when divided with 7 will never give remainder 5. Prove it
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do it by own because try try but dont cry
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Step-by-step explanation:
Let's put the integers in the format of a=7n-m, 0≤m<7
(7n-m)²= 7n(7n-2m)+m², since the first part is the factor of 7, m²/7 will give us the remainder:
- m=0 ⇒ rem=0
- m=1 ⇒ rem =1
- m=2 ⇒ rem=4
- m=3 ⇒ rem= 9-7= 2
- m=4 ⇒ rem= 16-14= 2
- m=5 ⇒ rem= 25-21= 4
- m=6 ⇒ rem= 36- 35= 1
So we only got 1, 2, 4 as possible remainder and 5 is not one of them
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