Q12/125) Assume x and y are integers, such that (x^2+1)=2y. Now prove that y is the sum of squares of two integers?

x has to be odd so let x= 2m + 1

x^2 + 1 = (2m+1)^2 + 1 = 4m^2 + 4m + 2

(x^2+1)/2 = y = 2m^2 + 2m + 1 = m^2 + (m+1)^2

proved

Q2/124) Show that in a sequences of 3 numbers one number is always divisible by 3

This can be proved by pigeon hole principle. As there are 3 numbers there are 3 remainders and the remainders can be 0 or 1 or 2 and no 2 remainder can be same if they where same then difference is divsible by 3 but difference cannot be be >2 so this is not possible.

So one of the remainders has to be zero.

Hence proved.

However this can be proved as below

Add 3 to all the terms starting from 1^st number and keep adding. Then we get all the terms to the right. Subtract 3 from 3^rd term and keep subtracting. We get all the numbers and none of them is divisible by 3. which is a contradiction. So one of the numbers has to be divisible by 3.

Q2/123) Find integer sided isosceles triangle whose area is integer

We have Pythagorean triplet that is integer length sides for a right angled triangle with hypotenuse c as

a= u^2-v^2
b= 2uv
and c = (u^2+v^2)

Now it may be noted that as b is even so ab/2 is integer.

Now if we double any of the base and hypotenuse as the two other sides we have area is integer

So the 3 sides are (u^2+v^2, u^2+v^2, 4uv) or (u^2+v^2, u^2+v^2, 2u^2-2v^2)

Q2/122) : find a , b, c,d such that 1/a = 1/b + 1/c + 1/d

We know that

1/n = 1/(n+1) + 1/(n(n+1))  .. (1)

Puting n = a we get

1/a = 1/(a+1) + 1/(a(a+1))

Now putting n = a(a+1)

We get

1/(a(a+1)) = 1/(a(a+1) + 1) + 1/((a(a+1))(a(a+1) + 1) =

= 1/(a^2 + a + 1) + 1/(a(a+1)(a^2+a+1))

Hence b= a+ 1

c= a^2 + a + 1 and d = abc is the solution

for example if a= 2 we have

1/2 = 1/3 + 1/6 ... using 1

1/6 = 1/7 + 1/42 using 1

hence 1/2 = 1/3 + 1/7+ 1/42 ( a= 2 , b= a+1 = 3, c= a^2 + a + 1 = 7 and d= abc = 42)

we can proceed indefinitely to any number of reciprocals. and for any starting reciprocal as well

Q2/121) The set of all integers n for which sqrt(n^2 + n) is an integer is.

a) the set {0,-1};
b) a finite set with at least three elements;
c) an infinite set;
d) none of these sets;

Proof:
sqrt(n^2 + n) = sqrt(n(n+1)) is inetger

n and n+1 are coprimes so either n= 0 or n = - 1 or n and n+1 both squares

n = x^2 and n+1 = y^2 => 1 = (x+y)(y-x) => x+y = 1 and y-x = 1 => x = 0 y =1 => n = 0

or x+y = -1 and y-x = -1 => x = 0 y =-1 => n = 0

so only solution 0 and -1 so ans is a)

Q12/120) Let a,b and c be the sides of a right angled triangle. Let theta be the smallest angle of this triangle.?

f 1/a, 1/b, 1/c are also the sides of a right angled triangle then show that Sin(theta) = (sqrt(5) - 1)/2;

Proof:
Let a < b < c


And theta (say t opposite to smaller side)

Sin t = a/c and cos t = b/c

Now 1/a > 1/b > 1/c and as it is right angled triangle we have

(1/a)^2 = (1/b)^2 + (1/c)^2

Or

(c/a)^2 = (c/b)^2 +1

(1/sin ^2 t) = (1/ cos^2 t ) +1

Or cos^2 t = sin ^2 t + sin ^2 t cos^2 t

Or 1- sin ^2t = sin ^2 t + sin ^2 t (1- sin ^2 t)
= 2 sin ^2 t – sin ^4 t
Or sin ^4 t – 3 sin ^2 t + 1 = 0

Sin^2 t = (3 +/- sqrt(9-4))/2 = (3 – sqrt(5))/2 , + cannot be taken as it shall be >1 not possible


So sin t = sqrt((3 – sqrt(5))/2) say sqrt(x) – sqrt(y) as there is sqrt(5) in square

So squaring we get

x+y – 2sqrt(ab) = (3 –sqrt(5))/2

so x+y = 3/2 and sqrt(xy) = sqrt(5)/4) or xy = 5/16

we can solve it as x = 5/4 and y= 1/4 or y = 1/4 and x = 5/4

so sin t = +/-(sqrt(5/4) - sqrt(1/4))

positive value to be taken as sint > 0 so sin t = sqrt(5/4) – sqrt(1/4) = (sqrt(5)-1)/2
 

Q12/119) Let a < b < c < d be four real numbers, such that all six pairwise sums ...? are distinct. The values of the smallest four pairwise sums are 1, 2, 3, and 4 respectively. What are the possible values of d?

Then the six pairwise sums

a+b, a+c, a+d, b+c, b+d, c+d

are all distinct.
there are 2 possibilities
case 1:
a+d < b+ c

Now, a+b = 1, a+c = 2, a+d = 3, b+c = 4.

∴ (a+c) - (a+b) = 2-1

∴ c-b = 1

∴ (b+c) + (c-b) = 4+1 ∴ 2c = 5 ∴ c = 5/2

∴ a+c = 2 gives a = 2-c = 2-(5/2) = -1/2

∴ a+d = 3 gives d = 3-a = 3-(-1/2) = 7/2

∴ d = 7/2. ........ Ans for case 1.
 case2 : b+ c< a + d
then we get a+b = 1, a+c = 2, b+c =3 a +d = 4 => 2a + b + c = 3 => a = 0 b = 1 c = 2 and d = 4

so d =4 is another solution for case 2

so d can be 7/2 or 4