prove that the square of any positive integer is of the form 4qor4q+1 for some integer
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mkrishnan:
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Concept :
Bring all the equations in the form of a = bq + r where r ≥ 0.
Let any positive integer be a.
Case 1 : When a = 4m
On squaring both the sides we get,
⇒ a² = (4 m)²
⇒ a² = 16 m² = 4(4 m²)
⇒ a² = 4q [ where q = 4 m² ]
Case 2 : When a = 4m + 1
On squaring both the sides we get,
⇒ a² = (4 m + 1)²
⇒ a² = 16 m² + 1 + 8 m
⇒ a² = 4(4 m² + 2 m) + 1
⇒a² = 4q + 1 [ where q = 4 m² + 2 m ]
Hence square of any positive integer is in form of 4q or 4q + 1.
Q.E.D
so you should take a =2m,a=2m+1 only two cases
your cases are not fulfil all possible
But questioner asked only 4q and 4q + 1
Dear refer to RD Sharma for solution. My answer is correct
If you feels its incorrect then you are free to report it.
if a is any positive integer is it only 2 possible 4q and 4q +1 ?
isnt a chance for a to be 4q +2 or 4q +3 ?
you should prove a^2 is in 2 form 4m or 4m +1
but you take a=4q and 4q +1
see the other answer that is perfect
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