Question

BRAINLEST FOR BEST ANSWER
use eyelids division lemme to show that the square of any positive integer cannot be of the form 5m+2 or 5m+3 for some integer m​

Answer 1

Answer:

Hi. Good morning . I will not give you full answer. Just the hint would be given to you. The reason being that i want you to practice yourself. I know you will not mark me brainliest if i don't give the answer but as i said it is more important for you to learn.

Now. Using the Euclid's divisions algorithm which says that

a=bq+r

Let a be any positive integer.

b=5. . r= 0,1,2,3,4.

Then . a= 5q+r

Now put the values of r and square it.. Represent the squared nber in the form of 5q,5q+1 or 5q+4. You will get your answer.

Hope you understand what i am saying

Answer 2

Here is your ans.,

Let there be any positive integer ' a '

such that b = 5


Acc. to Euclid 's Division Lemmma :

a = 5q + r
where r is less than 5


so if r = 1 ,
a^2 = ( 5q+1)^2

= 25q^2 + 1 + 10q
= 5( 5q^2+2q ) + 1
( where 5q^2 + 2q = m )

= 5m+ 1


If r = 2
a^2 = ( 5q+2)^2
= 25q^2 + 4 + 20q
= 5( 5q^2 +4q) +4
( where 5q^2 +4q= m )
= 5m+4


if r= 3
a^2=(5q+3)^2
=25q^2 +9+30q
= 25q^2 +30q + 5 + 4
= 5( 5q^2 + 6q + 1) + 4
( where 5q^2 + 6q + 1 = m )
= 5m+4

if r=4
a^2 = ( 5q + 4 )^2
= 25q^2 + 16 + 40q
= 25q^2 + 40q +15 + 1
= 5( 5q^2 + 8q + 3 ) + 1
( where 5q^2 + 8q + 3 =m )
= 5m+ 1


Hence proved , square of any positive integer cannot be of the form 5m+ 2 and 5m+3.



Hope it helps...


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