Question

Show that the square of any positive integer is of the form 5q,5q+1,5q+4 for one integer q by using exclid s division lemma

Answer 1

AnswEr :

Let us Consider that a & b are two positive integers.

$\:\bold{\underline{\underline{\sf{\pink{By,\: Using\; Euclid's\; Division\: lemma}}}}} :-$

$\implies$ a = bm + r & 0 < r < b

$\implies$ b = 5 So, r can be 0, 1, 2, 3 & 4

$\rule{150}2$

$\implies$ a = 5m + r

If r = 0

$\implies$ a = 5m

$\:\:\:\;\:\;\:\;\dag\small\bold{\underline{\underline{\sf{\red{Squaring}}}}} -$

$\implies\sf (5m^2)$

$\implies\sf 25m^2$

$\small\sf{Taking\: Common\: 5}$

$\implies\sf 5(5m^2)$

$\sf{So, \; 5q \:\;\;\;\;\;\;\:\:\:\;[Here,\; q \:=\: 5m^2]}$

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If r = 1

$\implies\sf 5m + 1$

$\:\:\:\;\:\;\:\;\dag\small\bold{\underline{\underline{\sf{\red{Squaring}}}}} -$

$\implies\sf (5m + 1)^2$

$\sf{Using\: Identity\: (a +b)^2\:=\;a^2 + b^2 + 2ab}$

$\implies\sf 25m^2 + 1 + 10m$

$\small\sf{Taking\: Common\: 5}$

$\implies\sf 5(5m^2 + 2m) +1$

$\sf{So, 5q + 1 \:\;\;\;\;\;\;\:\:\:\;[Here,\; q \:=\: 5m^2 + 2m]}$

$\rule{150}2$

If r = 2

$\implies\sf 5m + 2$

$\:\:\:\;\:\;\:\;\dag\small\bold{\underline{\underline{\sf{\red{Squaring}}}}} -$

$\implies\sf (5m +2)^2$

$\implies\sf 25m^2 + 4 + 20m$

$\small\sf{Taking\: Common\: 5}$

$\implies\sf 5(5m^2+4m) + 4$

$\sf{So, \; 5q + 4 \:\;\;\;\;\;\;\:\:\:\;[Here,\; q \:=\: 5m^2 + 4m]}$

$\rule{150}2$

If r = 3

$\implies\sf 5m + 3$

$\:\:\:\;\:\;\:\;\dag\small\bold{\underline{\underline{\sf{\red{Squaring}}}}} -$

$\implies\sf (5m + 3)^2$

$\implies\sf 25m^2 + 30m + 9$

$\implies\sf 25m^2 + 30m + 5 + 4$

$\small\sf{Taking\: Common\: 5}$

$\implies\sf 5(5m^2 + 6m + 1)+4$

$\sf{So, 5q + 4 \:\;\;\;\;\;\;\:\:\:\;[Here,\; q \:=\: 5m^2 + 6m + 1]}$

$\rule{150}2$

If r = 4

$\implies\sf 5m + 4$

$\:\:\:\;\:\;\:\;\dag\small\bold{\underline{\underline{\sf{\red{Squaring}}}}} -$

$\implies\sf (5m +4)^2$

$\implies\sf 25m^2 + 40m + 16$

$\implies\sf 25m^2 + 40m + 15 + 1$

$\implies\sf 5(5m^2 + 8m + 3) +1$

$\sf{So, 5q + 1 \:\;\;\;\;\;\;\:\:\:\;[Here,\; q \:=\: 5m^2 + 8m +3]}$

Hence, Square of any positive integer is of the form 5q, 5q+1, 5q+4 for some integer q.

Answer 2

♻️ANSWER♻️

Let positive integer a = 5m+ r , By division algorithm we know here 0 ≤ r < 5 , So

When r = 0

a = 5m

Squaring both side , we get

a2 = ( 5m)2

a2 = 5 ( 5m2)

a2 = 5q, where q = 5m2

When r = 1

a = 5m + 1

squaring both side , we get

a2 = ( 5m + 1)2

a2 = 25m2 + 1 + 10m

a2 = 5 ( 5m2 + 2m) + 1

a2 = 5q + 1 , where q = 5m2 + 2m

When r = 2

a = 5m + 2

Squaring both hand side , we get

a2 = ( 5m + 2)2

a2 = 25m2 + 5 + 20m

a2 = 5 ( 5m2 + 4m + 5)

a2 = 5q , Where q = 5m2 + 5m + 1

When r = 3

a = 5m + 3

Squaring both hand side , we get

a2 = ( 5m + 3)2

a2 = 25m2 + 9 + 30m

a2 = 25m2 + 30m + 10- 1

a2 = 5 ( 5m2 + 6m + 2) - 1

a2 = 5q -1 , where q = 5m2 + 6m + 2

Hence

Square of any positive integer is in form of 5q or 5q + 4. , where q is any integer.