Q13/010) Let p be a prime > 2, then 1+(1/2)+(1/3)+...+(1/p-1)=j/k, j,k>0 and integers. Show that j is divisible by p

Proof:

Because p is prime so for each m (not -1, not p-1 for these these are their own reciprocals ) there is a reciprocal q such that

mq = 1 mod p or mq = a(m)p + 1 ( we use a(m) for coefficient of m so on and a(m) = a(q))

so we have m = (a(q)p + 1)/q = a(q) p/q + 1/q

or 1/q = m – a(q) p/q

1/m = q – a(m) p/m

 

For each different m there is a different reciprocal.

 

This is for all except 1 and p-1

1/1 = 1

1/(p-1) = 1

 

Adding all the expressions on the left and right we get

1+(1/2)+(1/3)+...+(1/p-1)= ( 1+ 2 + …. + p-1)) + (sum (a(i)/i) p

= p(p-1)/2 + + (sum (a(i)/i) p

Which on adding has p in numerator and not on denominator

 

Q13/009)Find all pairs of positive integers x and y such that 2x+1 is divisible by y and at the same time 2y+1 is divisible by x.

We have

2x + 1 = ny ... 1
2y + 1 = mx .... 2 with m and n > 0. Without loass of generality let m >= n
from the above (1) and(2) we have n,y,m and x all are odd

Again from (1)
4x + 2 = 2ny = n(mx-1) as 2y = mx-1 from (2)
or 4x + 2 = nmx - n
or x(nm-4 ) = n+ 2
or x= (n+2)/(nm- 4)

now as x n + 2 >= nm - 4 or n(m-1) <= 6 and nm > 4
n =1 => m= 5 or 7

n =3 => m = 3

n= 1, m= 5 => x = 1 and y = 3

n =1,m =7 => x = 3, y = 7

n= 3 , m= 3 => x = 1 , y =1  

So solution set (1,1) , (1,3) (3,7)

Q13/008) Integrate sin^-1 (cos x) 0 <= x < = pi/2

we have sin^-1( cos x) = pi/2 - x as cos x = sin (pi/2 - x)

so integration of sin^-1 (cosx) is pi x/ 2 - x^2/2 + C

Q13/007) Let P = 2 * 4 * 6 * ... * 400 Let Q = 1 * 3 * 5 * ....* 399 Prove that (P-Q) is divisible by 401.

Solution

we know 

2 = - 399 mod 401

4= - 397 mod 401

6 = - 395 mod  401

.

.

400 = -1 mod 401

multiply all and rearrange all terms ( knowing that there are 200 - signs) we get the result

Q13/006)Let f(x) = (x-a) (x-b) (x+a) (x+b). Find all integers a and b such that f(x) = a^2 b^2 has exactly 3 distinct integer roots.

As it is a degree 4 polynomial it has 4 root so it must have a double toot and 2 single roots

f(x) = (x-a) (x-b) (x+a) (x+b) = a^2b^2
=> (x^2 - a^2)(x^2 - b^2) = a^2b^2
=> x^4 - (a^2 + b^2)x^2 + a^2b^2 = a^2b^2
=> x^2 [x^2 - (a^2 + b^2)] = 0
=> x = 0 or x = ±√(a^2 + b^2)

This has a double root 0 and 2 more roots when a,b are sides of a Pythagorean triangle or a or b ( but not both) = 0

Q13/005) If an integer sided right triangle has two prime sides, must the difference of some two sides equal one?

If x y z are a triple then

x^2+y^2 = z^2

=> x^2 = (z^2-y^2)= (z+y)(z-y)

if x is prime then x^2 cannot have 2 different factors

(z-y) and (z+y) being different so z –y has to be 1

Q13/004) Solve in positive integers x + y + z = 97 and (4x/5) + (5y/6) + (6z/7) = 82.

We are given

x + y + z = 97 …1

(4x/5) + (5y/6) + (6z/7) = 82. …(2)

Subtract (2) from (1) to get

x/5 + y/6 + z/7 = 15  … (3)

now as 5,6,7 ar pairwise co primes

x is a multiple of 5, y of 6 and z of 7

so x = 5a, y = 6b, z = 7c where a,b,c are integers

so 5a + 6b + 7c = 97 ..(4) from (1)

a+ b+c = 15   … (5) from (2)
multiply (5) by 5 and subtract from (4) to get

b + 2c = 22  … 6

subtract (5) from (6) to get

c = 7+ a

so solution set as c = a + 7 so b= 22-2(a+7) = 8- 2a from (6)

so (a, b = 8-2a, c= a+7) is the solution over integers

for positive integers a can be 1 or 2 or 3

a = 1,  b = 6, c = 8=> (x,y,z) = (5,36,56)

a = 2, b= 4, c= 9 =>(x,y,z) = (10,24,63)

a=3 , b= 2, c= 10=>(x,y,z) = (15,12,70)

Q13/003) Find the unique triple (x,y,z) of positive integers such that x < y < z and (1/x)-(1/xy)-(1/xyz)=19/97

(1/x)-(1/xy)-(1/xyz)=19/97

=> 1- 1/y – 1/yz = 19x/97

Or 1/y + 1/yz = (97-19x)/97 so 19 x < 97 or x <=5

Now as 97 is in denominator on RHS so z has to be multiple of 97

If the value of z is chosen to be < 97 for simplicity and no solution with higher z can be proved

x = 1 => 1/y + 1/yz = 78/97 no solution as y has to be 2 and 1/2 + 1/6 = 2/3 < 78/97

x = 2 => 1/y + 1/yz = 59/97 no solution as 1/3 + 1/12  = 5/12 < 59/97

x= 3 => 1/y+ /yz = 40/97 no solution as 1/4 + 1/20 = 3/10 < 40/97

x= 4 => 1/y + 1/yz = 21/97 no solution as 1/5 + 1/(5* 97) = 98/(5 * 97) < 21/97

x =5 => 1/y + 1/yz = 2/97

or 97z + 97 = 2yz

or y = 97(z+1)/(2z)

if y = 97m then m = (z+1)/2z. m > 1 is not possible and m = 1 => z = 1 but z > y so no solution

z has to odd and multiple of 97

z= 97 => y = 49

z = 97 *3 => y = 146 and no solution for z > 97

so z = 97 and y = 49 is solution

so x = 5, y = 49 and z = 97 is the solution

Q13/002) Prove that arranging odd positive integers in the following way: 1 3 + 5 7 + 9 + 11 13 + 15 + 17 + 19 21 + 23 + 25 + 27 + 29 ......................................… produces the sequence of the cubes of the integer more precisely the sum of each line is the cube of the quantity of the numbers

The nth line has got n numbers.

now the 1st number
1,3 ,7 ,13, 21 ..

1st order difference 2,4,6, 8

2nd order difference = 2,2,2,

as it is constant we have this equation is quadratic in n

so an^2 + bn + c
putting n =1 ,2 and 3 for 1st 3 terms

n =1 => 1 = a + b+ c ..1
n= 2 => 3 = 4a + 2b + c ..2
n =3 => 7 = 9a + 3b + c ...3

from (1) and (2) 3a + b = 2
from (2) and (3) 5a + b = 4

so a =1 and b= -1 and hence c = 1

so term =n^2 - n + 1

now in the nth row 1st term = n^2 - n + 1 and common difference = 2 and n terms

so last term = 2n^2 - n + 1 + 2(n-1) = n^2 +n - 1

so the average = 1/2(n^2 - n + 1 + n^2 + n -1) = n^2
as there are n terms sum = n^3

Q13/001) If x^2+x+1=0,then find the value of [x^3+1/x^3]^3

x^2 + x + 1 = 0
so x^2 + 1 = - x
(x + 1/x) = -1


cube both sides

(x^3 + 1/x^3) + 3(x+ 1/x) = - 1
or (x^3 + 1/x^3) - 3 = - 1
or (x^3 + 1/x^3) = 2 hence
(x^3 + 1/x^3)^3 = 8