2025/007) If 2n+1 and 3n+1 are both perfect squares, then how could we prove that 40|n?

For n this to be divisible by 40 we need to show that it is divisible by 8 and 5 as 40 = 8 * 5 and 8 and 5 are co-primes  

 First let us prove that 8 divides n

Clearly 2n+1 is odd so this has to be square of odd number say 2k+1

Now $(2k+1)^2 = 4k^2 + 4k +1=4k(k+1)+1 = 4m+1$ when $m= k(k+1)$

or $2n+1 = 8m+1$ or $n= 4m$

so n is even and hence 3n+1 is odd so  square of odd number say 2p+1

$3n+1 = (2p+1)^2 = 4p^2 + 4p +1 = 4p(p+1) +1$

or $3n = 4p(p+1)$ as p(p+1) is even it can be written as 2q

as $3n+1 = 8q+1$5

or $3n = 8q$

So 8 divides 3n and as 8 and 3 are copimes so 8 divides n 

now let us prove that 5 divides n

let us find what are the numbers that can be remainder of $x^2$ when divided by 5

we have x of the form $ 5q+k$ where k is one of $0,\pm 1,\pm 2$

$x^2= 25q^2 + 10kq +k^2$

$= 5(5q^2 + 2k) + k^2$

so remainder when $x^2$ divided by by5 is 0 or 1 or 4

now let us find n for which 2n+1 give a remainder 0 or 1 or 4 else it cannot be a square.

$n = 5k$ gives $2n + 1 = 10 k  + 1 = 5 * 2k + 1$ remainder is     1 which is a valid candidate

$n = 5k+ 1$ gives  $2n + 1 = 10 k  +3 = 5 * 2k +3$ remainder is 3 which is a not a valid candidate

$n = 5k+ 2$ gives  $2n + 1 = 10 k  +5 = 5 * (2k +1)$ remainder is 0 which is a valid candidate

 $n = 5k+ 3 gives  $2n + 1 = 10 k  +7 = 5 * (2k +1)+2 $ remainder is 2 which is a not a valid candidate

$n = 5k+ 4$ gives  $2n + 1 = 10 k  +9 = 5 * (2k +1)+ 4 $ remainder is 4 which is a valid candidate

So valid candidates for 2n+1  being perfect square is 5k, 5k+2, 5k+ 4

and we need to check for these one 3n+1 is valid

we can apply above procedure to see  the$ n= 5k$ 3n+1 leaves remainder 1 which is valid and for $n=5k+2$ remainder is 2 and for $n=5k+4$ remainder is 3 both of which are invalid

so n is multiple of 5.

as n is divisible by 5 and 8 so by 40

Q13/127) find integers in positive solutions to xyz + 20xy + yz + 5zx + 100x + 20y + 5z = 1913

Because we have xyz, xy, yz, xz, x, y, z terms in the left we should add some d and factor if possible

(x+a)(y+b)(z+c) – d

= xyz + ayz + bzx + cxy + bcx + abz + acy + abc – d

Comparing with given expression we get a = 1,  b = 5, c = 20

So we get (x+1)(y+5)(z+20) = xyz+ yz +20xy + 5xyz + 100 x + 5z + 20y + 100

Comparing with coefficient we get d = 100

So (x+1)(y+5)(z+20) = 1913 + 100 = 2013 = 3 * 11 * 61

So x = 2, y = 6, z= 41

2011/041) factoring by diminishing power $6bc - 9c² - 12cd - 8be + 12ce +16de$

one of the factoring that I saw in the net was factor

$6bc - 9c² - 12cd - 8be + 12ce +16de$

...the "clue" you need to recognize is that the coefficients of the first three terms have a common factor...and the coefficients of the last three terms have a common factor...and, the ratios are constant...so, group accordingly...

$(6bc - 9c^2 - 12cd) - (8be - 12ce - 16de) = 3c(2b - 3c - 4d) - 4e(2b - 3c - 4d) =$
$(2b - 3c - 4d)(3c - 4e)$

the above is correct and it was luck that grouping was there but what is luck is not there
then see that highest power of c is 2 and keep them in descending order

$ 6bc - 9c² - 12cd - 8be + 12ce +16de$
$= - 9c² + 6bc - 12cd + 12ce - 8be + 16de$
now you can factor as quadratic in c eliminating the luck factor
$= - 9c² - + 6c( b - 2d + 2e) -8e(b- 2d)$
$= - 9c² - + 6c(( b - 2d) + 2e)) -8e(b- 2d)$
letting b - 2d = a we get
$= - 9c^2 + 6c(a + 2e) - 8ea$
$= - 9c^2 + 6ca + 12ec - 8ea$
$= - 3c(3c - 2a) + 4e(3c-2a)$
$= (3c-2a)(4e - 3c)$
$= (3c-2b+4d)(4e-3c)$
which is same as 1st one

I do not mean to say that 2nd one is preferable to 1st but 2nd one can be used when 1st one does not work

2011/013) factorization using short cuts.

1) (a+b+c)^3 - a^3-b^3-c^3

we know that (a+b+c)^3 - a^3 is divisible by (a+b+c) – a that is b+ c

and b^3+c^3 by (b+c)

hence b+c is a factor

by symmetry (a+b) and (c+a) are also factors

2) so it is (a+b+c)^3 - a^3-b^3-c^3 = m(a+b)(b+c)(c+a)

now as LHS does not contain a^3 or b^3 or c^3 (as they cancel out) so m has to be a constant

putting a =1 b = 1 and c = 1 we get LHS = 24 and RHS = 8m or m = 3

hence (a+b+c)^3 - a^3-b^3-c^3 = 3(a+b)(b+c)(c+a)

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)

2010/051) factor: x^2 - y^2 - z^2 -2yz + x + y + z

= x^2-(y^2+z^2+2yz) + (x+y+z)
= x^2-(y+z)^2 + (x+y+z)
= (x+y+z)(x-y-z) + (x+y+z)
= (x+y+z)(x-y-z+1)

2010/034) Factorise x^4+ x^2+1

x^3 and x are missing so we can try to do it by difference of squares adding and subtracting x^2 that is middle term

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

2010/030) factor a^2(b+c) + b^2(a+c) + c^2(a+b) + 2abc

a^2(b+c) + b^2(a+c) + c^2(a+b) + 2abc
= a^2(b+c) + (b^2a + b^2 c + ac^2 + bc^2 + 2abc)
= a^2(b+c) + a(b^2+c^2 + 2ab) + (b^2 c + bc^2)
= a^2(b+c) + a(b+c)^2 + bc(b+c)
= (b+c)(a^2+ ab + ac + bc)
= (b+c)(a+b)(a+c)

2010/016) Let 1, a1, ... ,a6 denote the distinct roots of x^7 - 1.

Let 1, a1, ... ,a6 denote the distinct roots of x^7 - 1. Then the product (1 - a1)(1 - a2)(1 - a3) ...(1 - a6) is

is
a) 0;
b) 1;
c) 6;
d) 7;
Kindly explain...
x^7- 1 = (x-1)(x-a1)(x-a2)(x-a3)(x-a4)(x-a5)(x-a6… by definition of roots

dividing by (x-1) on both sides

x^6+x^5+x^4+x^3+x^2+x+1 = (x-a1)(x-a2)(x-a3)(x-a4)(x-a5)(x-a6)

putting x= 1 on both sides we get 7 = (1-a1)(1-a2)(1-a3)(1-a4)(1-a5)(1-a6)

hence d

2009/032) factor f(x) = x^3-3x^2+4

by taking factors of 4 that is 1,-1.2,-2,4,-4 we see that

f(-1) = 0

so (x+1) is a factor

so x^3-3x^2+4
= x^3+x^2-4(x^2-1)
= x^2(x+1)-4(x+1)(x-1)
= (x+1)(x^2-4(x-1)
= (x+1)(x^2-4x+4)
= (x+1)(x-2)^2

2009/030) Factorize the polynomial 2{(bc)^2+(ca)^2+(ab)^2} - {a^4+b^4+c^4}

Factorize the polynomial
2{(bc)^2+(ca)^2+(ab)^2} - {a^4+b^4+c^4}?

We need to factorize
{a^4+b^4+c^4} – 2(bc)^2 - 2(ca)^2 - 2(ab)^2 and multiply by -1
Now
{a^4+b^4+c^4} – 2(bc)^2 - 2(ca)^2 - 2(ab)^2
= a^4 – 2a^2(b^2+c^2) + b^4+c^4-2ab^2c^2
= a^4 – 2a^2(b^2+c^2) + b^4+c^4+2b^2c^2- 4b^2c^2
= a^4 – 2a^2(b^2+c^2) + (b^2+c^2)^2- 4b^2c^2
= (a^2-b^2-c^2)^2 – (2bc)^2
= (b^2+c^2 – a^2)^2- (2bc)^2
= (b^2+c^2-a^2+2bc)(b^2+c^2-a^2-2bc)
= (b^2+c^2+2bc-a^2)(b^2+c^2-2bc-a^2)
= ((b+c)^2-a^2)((b-c)^2 – a^2)
= (b+c+a)(b+c-a)(b-c+a)(b-c-a)

So given expression
2{(bc)^2+(ca)^2+(ab)^2} - {a^4+b^4+c^4} = - (b+c+a)(b+c-a)(b-c+a)(b-c-a)
= (a+b+c))b+c-a)(a+b-c)(a+c-b) multiplying last term by -1 to get symmetrical form

2009/024) Resolve into factors X^4-3x+20

if this can be factored with real coefficients then it can be as product of 2 quadratic
we shall get
(x^2+ax+b)(x^2+cx+d)

now coeffcient of x^3 = 0 so a +c = 0 or c = -a

so we get(x^2+ax+b)(x^2-ax+d)
= x^4 + (d + b - a^2) x^2 + x(ad-ab) + bd

comparing coefficients

a^2 = b + d
bd = 20
so ad -ab = - 3
by trial and error we see that a = 3 , b= 5 and d= 4 (this is using 20 = 1 * 20 = 2(*10 - 4 * 5 outof which only 4 +5 = 9 we get perfect square) satisfies
this can be factored with rational coefficient as
(x^2+3x+5)(x^2-3x+4)
and not further this can be checked by quadaric factorisation

hence X^4-3x+20 = (x^2+3x+5)(x^2-3x+4)

2009/021) Factorize the polynomial x^4 + 4 with real coefficient?

x^4 + 4
= x^4 + 4x^2 + 4 - 4x^2
= (x^2 + 2)^2 - (2x)^2
= (x^2 + 2x + 2) (x^2 - 2x + 2)

2009/003) factor (x-y)^5+(y-z)^5+(z-x)^5

if a + b+ c = 0

then (a+b)^5 = -c ^5

so a^5 + 5 a^4 b + 10 a^3 b^2 + 10 a^2 b^3 + 5 a^4 b + b^5 = - c^5

so a^5 + b^5 +c^5 = - (5 a^4 b + 10 a^3 b^2 + 10 a^2 b^3 + 5 a^4 b)
= -5 ab(a^3 + 2 a^2b + 2 a b^2 + b^3)
= - 5 ab((a+b)^3 - (a^2 b + ab^2)
= - 5ab((a+b)3 - ab(a+b))
= - 5ab(a+b)((a+b)^2 - ab)
= 5abc(c^2-ab) as a+b = - c

as (x-y)+ (y-z) + (z-x) = 0

we get 5(x-y)(y-z)(z-x)((z-x)^2 - (x-y)(y-z))
= 5(x-y)(y-z)(z-x)(z^2 + x^2- 2xz -xy + xz +-y^2 -xy)
= 5(x-y)(y-z)(z-x)(x^2+y^2+z^2 - xy - yz - zx)