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Sia ? 6 years, 4 months ago
Since x and y are odd positive integers, Then it should be in the form of 2x+1 (where x is a positive integer)
Let x = 2m + 1 and y = 2n + 1 ( where m and n positive integers)
Now x2 + y2
= (2m + 1)2 + (2n + 1)2
= 4m2 + 4m + 1 + 4n2 + 4n + 1
= 4(m2 + n2 + m + n) + 2 ....... (1)
from (1) we get
x2 + y2 = 2 {m2 + n2 + m + n) + 1} = 2t (t = 2(m2 + n2 + m + n) + 1 is a positive integer
Therefore it is clear that {tex}x^2+y^2{/tex} is an even number but not divisible by 4.
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Sia ? 6 years, 4 months ago
Let P (x1, y1) Q(x2, y2) and R(x3, y3) be the points which divide the line segment AB into four equal parts.
Then, P divides AB in the ratio 1 : 3 internally.
{tex}x=\frac{mx_2+nx_1}{m+n}{/tex}
{tex}\therefore x _ { 1 } = \frac { ( 1 ) ( 2 ) + ( 3 ) ( - 2 ) } { 1 + 3 }{/tex}
{tex}= \frac { 2 - 6 } { 4 } = - \frac { 4 } { 4 } = - 1{/tex}
{tex}y=\frac{my_2+ny_1}{m+n}{/tex}
{tex}y _ { 1 } = \frac { ( 1 ) ( 8 ) + ( 3 ) ( 2 ) } { 1 + 3 }{/tex}
{tex}= \frac { 8 + 6 } { 4 } = \frac { 14 } { 4 } = \frac { 7 } { 2 }{/tex}
So, {tex}\mathrm { P } \rightarrow \left( - 1 , \frac { 7 } { 2 } \right){/tex}
Also, Q divides AB in the ratio 1 : 1 i.e.
Q is the mid point of AB
{tex}x _ { 2 } = \frac { - 2 + 2 } { 2 } = 0{/tex}
{tex}y _ { 2 } = \frac { 2 + 8 } { 2 } = \frac { 10 } { 2 } = 5{/tex}
So, {tex}Q \rightarrow ( 0,5 ){/tex}
and, R divides AB in the ratio 3 : 1
{tex}\therefore x _ { 2 } = \frac { ( 3 ) ( 2 ) + ( 1 ) ( - 2 ) } { 3 + 1 }{/tex}
{tex}= \frac { 6 - 2 } { 4 } = \frac { 4 } { 4 } = 1{/tex}
{tex}y _ { 3 } = \frac { ( 3 ) ( 8 ) + ( 1 ) ( 2 ) } { 3 + 1 }{/tex}
{tex}= \frac { 24 + 2 } { 4 } = \frac { 26 } { 4 } = \frac { 13 } { 2 }{/tex}
So, {tex}\mathrm { R } \rightarrow \left( 1 , \frac { 13 } { 2 } \right){/tex}
Posted by Fahad Mallick 7 years, 3 months ago
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Sia ? 6 years, 4 months ago
Let n be any positive integer.
By Euclid's division lemma, n = 5q + r, 0{tex}\leqslant{/tex}r < 5
n = 5q,5q + 1,5q + 2 ,5q 4- 3 or 5q + 4, where q{tex}\in \omega{/tex}
Now we find the square of n
If n=5q then (5q)2 = 25q2= 5(5q2) = 5m
If n=5q+1then n2= (5q + 1 )2 = 25q2 + 10q + 1 = 5m + 1
If n=5q+2 then n2 = (5q + 2)2 = 25q2 + 20q + 4 = 5m + 4
If n=5q+4 then n2=(5q + 3)2=25q2+30q+9=5m + 1
Thus square of any positive integer is in the form of 5m,5m+1 or 5m+4, hence cannot be of the form 5m + 2 or 5m + 3.
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