Question

Using Euclid’s Division Lemma, show that square of any positive integer is either of the form

4q, 4q+1 for some integer q.​

Answer 1

Answer:

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Answer 2

By Euclid's division algorithm, a = bq + r

Where a,b,q,r are non negative integer and 0 ≤ r < b.

On putting b = 4

We get,

$\sf\implies{a = 4q + r \: \: \: ....(i)}$

When r = 0, a = 4q

Squaring both sides

We get,

$\sf\implies{a {}^{2} = (4q) {}^{2} } \: \: \\ \sf\implies{a {}^{2} = 4(4q) {}^{2} }$

= 4m is perfect square for m

= 4q²

When r = 1, a = 4q + 1

Squaring both sides

We get,

$\sf\implies{a {}^{2} = (4q + 1) {}^{2} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf\implies{ = (4q) {}^{2} + (1) {}^{2} + 2(4q)} \\ \sf\implies{ = 4(4q {}^{2} + 2q) + 1 } \: \: \: \: \: \: \: \:$

= 4m + 1 is perfect square for m = 4q² + 2q

When r = 2, a = 4q + 2

Squaring both sides

We get,

$\sf\implies{a {}^{2} = (4q + 2) {}^{2} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf\implies{a {}^{2} = (4q) {}^{2} + (2) {}^{2} + 2(4q)(2) } \\ \sf\implies{a {}^{2} = 4(4q {}^{2} + 4q + 1) } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

= 4m is perfect square for m = 4q² + 4q + 1

When r = 3, a = 4q + 3

Squaring both sides

We get,

$\sf\implies{a {}^{2} = (4q + 3) {}^{2} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf\implies{ (4q) {}^{2} + (3) {}^{2} + 2(4q)(3) } \\ \sf\implies{16q {}^{2} + 9 + 24q } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf\implies{16q {}^{2} + 24q + 8 + 1} \: \: \: \: \: \: \: \\ \sf\implies{4(4q {}^{2}6q + 2) + 1 } \: \: \: \: \: \: \: \: \: \: \:$

= 4m + 1 is perfect square for some of value of m.

Hence, the square on any integer is either of the form 4q or 4q + 1 for some integer q.

$\huge\sf\underline\pink{Hence \: Proved \: ! }$