Question

Use Euclid’s Lemma to show that square of any positive integer is of form 4 m or 4m+1 for some integer m.​

Answer 1

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Let positive integer a = 4m + r ,

By division algorithm we know here 0 ≤ r < 4 ,

So

When r = 0

a = 4m 

Squaring both side , we get

a^2 = ( 4m )^2

a^2 = 4 ( 4m^2)

a^2 = 4 q , where q = 4m^2

When r = 1

a = 4m + 1

squaring both side , we get

a^2 = ( 4m + 1)^2

a^2 = 16m^2 + 1 + 8m 

a^2 = 4 ( 4m^2 + 2m ) + 1 

a^2 = 4q + 1 , where q = 4m^2 + 2m

When r = 2

a = 4m + 2 

Squaring both hand side , we get

a^2 = ( 4m + 2 )^2

a^2 = 16m^2 + 4 + 16m 

a^2 = 4 ( 4m^2 + 4m + 1 )

a^2 = 4q , Where q = 4m^2 + 4m + 1

When r = 3 

a = 4m + 3

Squaring both hand side , we get

a^2 = ( 4m + 3)^2

a^2 = 16m^2 + 9 + 24m 

a^2 = 16m^2 + 24m + 8 + 1

a^2 = 4 ( 4m^2 + 6m + 2) + 1

a^2 = 4q + 1 , where q = 4m^2 + 6m + 2 

Hence 

Square of any positive integer is in form of 4q or 4q + 1 , where q is any integer.

Answer 2

$\huge\red{\underline{{\boxed{\textbf{QUESTION}}}}}$

Use Euclid’s Lemma to show that square of any positive integer is of form 4 m or 4m+1 for some integer m.

$\huge\red{\underline{{\boxed{\textbf{ANSWER}}}}}$

Let positive integer a = 4m + r

By division algorithm,

we know here 0 ≤ r < 4 , So

When r = 0

a = 4m

Squaring both side , we get

a2 = ( 4m )2

a2 = 4 ( 4m2 )

a2 = 4 q , where q = 4m2

When r = 1

a = 4m + 1

squaring both side , we get

a2 = ( 4m + 1 )2

a2 = 16m2 + 1 + 8m

a2 = 4 ( 4m2 + 2m ) + 1

a2 = 4q + 1 , where q = 4m2 + 2m

When r = 2

a = 4m + 2

Squaring both hand side , we get

a2 = ( 4m + 2 )2

a2 = 16m2 + 4 + 16m

a2 = 4 ( 4m2 + 4m + 1 )

a2 = 4q , Where q = 4m2 + 4m + 1

When r = 3

a = 4m + 3

Squaring both hand side , we get

a2 = ( 4m + 3 )2

a2 = 16m2 + 9 + 24m

a2 = 16m2 + 24m + 8 + 1

a2 = 4 ( 4m2 + 6m + 2 ) + 1

a2 = 4q + 1 , where q = 4m2 + 6m + 2

Hence,

Square of any positive integer is in form of 4m or 4m + 1 , where q is any integer . ( Hence proved )

@HarshPratapSingh