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Let m be a natural number …

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Let m be a natural number for how many values of m 4m+1 is a perfect square

Posted by Akshita Gupta 6 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 6 months ago

Let a be any positive integer.
Applying Euclid’s division lemma with divisor = 2, we get
{tex}\begin{array}{l}a=2q+r\;\;\;\;\;\;\;\;\;0\leq r<2\\So\;r=0,1\\\end{array}{/tex}
When r = 0,
a = 2q
So a²= (2q)² = 4q² = 4m--------(1) ( where m = q², which is an integer)
When r = 1
Then a= 2q+1
a²= (2q + 1) ² = 4q² + 4q + 1 = 4(q² + q ) +1 = 4m+1 --------(2) (where m = q² + q, which is an integer)
From (1) and (2) We Can conclude that
The square of any positive integer is of the form 4m or 4m +1 for some integer m.

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Posted by Kanika . 1 month ago

CBSE > Class 10 > Mathematics