2011/010) Find integral solution of (1-i)^n = 2^n

We have 1-i = (sqrt(2) cis (-pi/4)

So (1-i)^n = 2^(n/2) cis (-npi/4) = 2^n

The modulo of LHS = 2^(n/2) and rhs = 2^n and both are same

If 2^(n/2) = 2^n or n = 0

Then = 2^(n/2) cis (-npi/4) = 1 = RHS

So n = 0 is the only solution

2011/009) one problem in AP

question

The fourth power of the common difference of an AP with integer entries is added to the product of any four consecutive terms of it. Prove that the resulting sum is the square of an integer.

proof

With out loss of generality we can take the 1st term of the 4 consecutive terms to be a and let the difference be t

a and t are integers

We have the 4 terms = a, a+t, a+2t, a+ 3t and 4th power of difference is t^4

Now a(a+t)(a+2t)(a+3t) + t^4

= (a(a+3t))((a+t)(a+2t)) + t^4
= (a^2+3ta)(a^3+3ta + 2t^2) + t^4

Letting a^2 + 3ta = p we get

= p(p+2 t^2) + t^ 4= (p^2+ 2pt^2+t^4) = (p+t^2)^2 = (a^2+3at+t^2)^2

Which is square of integer as a^2+3at+t^2 is integer

2011/008) factor (a+b+c) ^3- a^3-b^3-c^3

we realize that

(a+b+c)^3-a^3 has a factor (b+c) and b^3 + c^3 has a factor b+c

so we proceed by combining

(a+b+c) ^3- a^3-b^3-c^3
= ((a+b+c) ^3- a^3)-(b^3+c^3)
= (b+c)((a+b+c)^2 + a(a+b+c) + a^2) - (b+c)(b^2+c^2-bc)
= (b+c)((a+b+c)^2+ a(a+b+c) + a^2 - b^2 - c^2 + bc)
= (b+c)(a^2+b^2+c^2+2ab + 2ac + 2bc a^2+ab+ac + a^2 - b^2 - c^2 + bc)
= (b+c)(3a^2 + 3ab + 3bc + 3ca)
= 3(b+c)(a^2+ab+bc+ca)
= 3(b+c)(a+c)(a+b)

2011/007) the total number of integer pairs (x, y)

the total number of integer pairs (x, y) satisfying the equation x + y = xy is
a) 0
b) 1
c) 2
d) 3
None of these

we have x+y- xy = 0
subtract 1 from each side

x+y - xy - 1 = - 1
x(1-y) + (y-1) = -1

or (x-1)(1-y) = - 1
so x-1 = 1 and 1-y = -1 => x = 2 and y= 2

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

so 2 pairs

2011/006) If x > 1, y > 1, z > 1 are in G.P.

If x > 1, y > 1, z > 1 are in G.P., then 1/(1+In x), 1/(1+In y),1/(1+ In z) are in :
(A) A.P.
(B) H.P.
(C) G.P.
(D) none of these

(this is objective and should not take more than a minute)

ans is B

reason
x > 1, y > 1, z > 1 are in G.P. so ln x, ln y, ln z are in AP and >0

so 1 + ln x, and 1+ in y and 1 + ln z in AP and hence the result

2011/005) if cos A + cos B + cos C = 0 = sin A + sin B + sin C

prove that

cos 3A + cos 3B + cos 3C = 3 cos( A+ B+ C)

proof:

we have

cos A + cos B + cos C = 0 ...1

sin A + sin B + sin C = 0 ....2

multiply 2nd by i and add

(cos A + i sin A ) + ( cos B + i sin B) + (cos C + i sin C) = 0

or e^(iA) + e^(iB) + e^(iC)= 0 as e^ix = cos x+ i sin x

and as if (x+y+z) = 0 then x^3+y^3 + z^3 = 3xyz

so e^(i3A) + e^(i3B ) + e^(i3C) = 3 e^i(A+B+C)

so ( cos 3A + i sin 3 A) + (cos 3B + i sin 3B ) + ( cos 3C + i sin 3C) = 3 cos (A+B+C) + 3i sin (A+B+ C)

or (cos 3A + cos 3 B + cos 3C) + i( sin 3A + sin 3B + sin 3C) = 3 cos (A+B+C) + 3i sin (A+B+ C)

equating real and imaginary parts on both sides we get
cos 3A + cos 3 B + cos 3C = 3 cos (A+B+C)
sin 3A + sin 3B + sin 3C = 3 sin (A+B+ C)

2011/004) Using mathematical induction prove that tan ^-1 (1/3) + tan ^-1 (1/7) + … tan ^-1 (1/(n^2+n+ 1) = tan ^-1(n/(n+2))

in mathematical induction there are 2 steps

number one base step

for n =1 LHS = tan ^-1 (1/3) and RHS = tan ^-1 (1/(1+2)) = tan ^-1 (1/3)

hence true

then the induction step

let it be true for n = k that is

tan ^-1 (1/3) + tan ^-1 (1/7) + … tan ^-1 (1/(k^2+k+ 1) = tan ^-1(k/(k+2))

for n = k + 1 we have
tan ^-1 (1/3) + tan ^-1 (1/7) + … tan ^-1 (1/(k^2+k+ 1) + tan ^-1(1/(k+1)^2 + k+ 1+ 1)
= tan ^-1(1/(k+2)) + + tan ^-1(1/(k+1)^2 + k+ 1+ 1)
= tan ^-1(1/(k+2)) + + tan ^-1(1/(k^2+3k + 3)

The RHS = tan ^-1((k+1)/(k+3))


As tan (A+ B) = (tan A + tan B)/(1- tan A tan B)

So A + B = tan ^-1(tan A + tan B)/(1- tan A tan B))

So tan ^-1(k/(k+2)) + tan ^-1(1/(k^2+3k + 3)
= tan ^-1 ( k/(k+2) + 1/(k^2+3k + 3)/(1-k/(k+2)1/(k^3+3k+ 3)
= tan ^-1(( k^3+3k^2+3k + k+ 2)/((k+2)(k^2+3k+3)-k)
= tan ^-1((k^3+3k^2 +4k + 2) /(k^3+ 5k^2 + 8 k + 6))

We have k^3+3k^2+ 4k + 2 = (k^3+ 3k^2 + 3 k + 1) + (k+1) = (k+1)^3 + k+ 1= (k+1)(k^2 + 2k + 2)

(k^3+ 5k^2 + 8 k + 6) = k^3 + 3k^2 + 2k^2 + 8k + 6
= k^2(k+3) + 2( k^2 + 4k + 3) = k^2(k+3) + 2( k+1) (k+ 3) = (k+ 3)( k^2 + 2k + 2)

So tan ^-1((k^3+3k^2 +4k + 2) /(k^3+ 5k^2 + 8 k + 6)) = tan ^-1 ((k+1)/(k+ 3)

= RHS

So induction step is proved

hence proved

2011/003) If 1 , w , w^2 are cube root of unity

If 1 , w , w^2 are cube root of unity show that (a+bw+cw^2)/(b+cw+aw^2) + (a+bw+cw^2)/(c+aw+bw^2) = -1

Proof:

As 1 w and w^2 are cube root of 1 we have

1+w+w^2 = 0 … 1

w^3 = 1…2


now
(a+bw+cw^2) = aw^3 + bw+ cw^2 = bw+cw^2 + aw^3= w( b+ cw + aw^2)

Hence (a+bw+cw^2)/(b+cw^2+aw^2) = w

Further
As (a+bw+cw^2) = a + bw + c/w (as w^2 = 1/w) = 1/w(aw+bw^2+c)

So (a+bw+cw^2)/ (c+ aw+bw^2 ) = 1/w = w^2

Hence (a+bw+cw^2)/(b+cw^2+aw^2) + (a+bw+cw^2)/ (c+ aw+bw^2 ) = w + w^2 = - 1 (from 1)

2011/002) product of sum of squares is sum of squares

Prove that product of two numbers, each of which can be expressed as sum of two squares, can itself be expressed as sum of two squares.

Let 1st number be a^2+b^2 and second be c^2 + d^2
one way

(a^2+b^2)(c^2+d^2)
= (a^2c^2 + b^2 d^2 + a^2d^2 + b^2 c^2)
= (a^2c^2 + b^2 d^2 + 2 abcd + a^2d^2 + b^2 c^2- 2abcd)
= (ac+bd)^2 + ( ad - bc)^2

also
(a^2+b^2)(c^2+d^2)
= (a^2c^2 + b^2 d^2 + a^2d^2 + b^2 c^2)
= (a^2c^2 + b^2 d^2 - 2 abcd + a^2d^2 + b^2 c^2+ 2abcd)
= (ac-bd)^2 + ( ad + bc)^2

so we can do in two different ways

further we can prove using complex numbers

(a^2+b^2)(c^2+d^2) = | a + ib|^2 |c + id|^2
= | (a + ib)( c+ id)| ^ 2
= | (ac - bd) + (ad +bc) i | ^2
= (ac-bd)^2 + (ad + bc)^2

and taking (a^2+b^2)(c^2+d^2) = | a + ib|^2 |c - id|^2

we get (ac+bd)^2 + (ad - bc)^2

2011/001) a problem in inequality

Let a,b,c,d be positive real numbers such that abcd = 1. Show that,?
(1 + a)(1 + b)(1 + c)(1 + d) >= 16

proof:

as a is real

(1+a) = (1-sqrt(a))^2 + 2 sqrt(a) or

1+ a >= 2 sqrt(a) ( we can also show it by AM GM inequality)

similarly

(1+b) > = 2 sqrt(b)

(1+c) > = 2 sqrt(c)

(1+d) > = 2 sqrt(d)

by multiplying

(1+a)(1+b)(1+c)(1+d)>= 16 sqrt(abcd) or > 16 as abcd = 1