Solution
We are given
$x+4\sqrt{xy} -2 \sqrt{x} - 4 \sqrt{y} + 4y =3$
Adding 1 to both sides
$x+4\sqrt{xy} -2 \sqrt{x} - 4 \sqrt{y} + 4y + 1=4$
Or $ (\sqrt{x} + 2\sqrt{y} -1)^2 = 4$ or $ (\sqrt{x} + 2\sqrt{y} -1) = 2$ as both square roots are positive
So $ (\sqrt{x} + 2\sqrt{y}) = 3$
So $\frac{\sqrt{x} + 2\sqrt{y} + 2014}{ 4- \sqrt{x} - 2\sqrt{y}}= \frac{3 + 2014}{ 4- 3} = 2017$
Because a,b ,c ,d are roots of the equation $x^4−7x^3+3x^2−21x+1=0$
We have by vieta's formula
$a+b+c+d = 7\cdots(1)$
Let
$f(x) = x^4−7x^3+3x^2−21x+1=(x-a)(x-b)(x-c)(x-d)\cdots(2)$
From (1) we have
$a+b+c = 7-d\cdots(3)$
$b+c+d = 7 -a\cdots(4)$
$c+d+a = 7 -b\cdots(5)$
$d+a +b = 7-c\cdots(6)$
From (3), (4), (5),(6) we have
$(a+b+c)(b+c+d)(c+d+a)(d+a+b)= (7-d)(7-a)(7- b)(7-c) = f(7)\cdots(7)$ from (2)
As $f(x) = x^4−7x^3+3x^2−21x+1$
So $f(7) = 7^4 - 7 * 7^3 +3 *7^2 - 21 * 7 + 1 = 1$
From (7) and above we have
$(a+b+c)(b+c+d)(c+d+a)(d+a+b) = 1$
Let $f(n) = n^4+6n^3+11n^2+6n+1$
This does not have a change of sign so there is no positive root
So it can have -ve root if it has real root it has to be -1
Checking using the rational root theorem $(f(-1) = 1 - 6 + 11 -6 +1 = 1$
So -1 is not a root
So it has to be product of 2 quadratic polynomials
This is a quartic polynomal. And the coefficients are symmetric.
So if n is a root then $\frac{1}{n}$ is a root. So the coeffcient of$x^2$ and constant should same in both polynomials. However I shall continue as below
$f(n) = n^4+6n^3+11n^2+6n+1$
$= (n^4+1) +11n^2+6(n^3+n)$ reordering the terms
$= (n^2+1)^2 -2n^2 +11n^2+6n(n^2+1)$ putting $n^4+1$ in terms of $(n^2+1)$
$=(n^2+1)^2 + 6n(n^2+1) + 9n^2$
$= (n^2+1)^2+ 2(3n)(n^2+1) + (3n)^2$ geting expressing in $a^2+2ab+b^2$ form
$=(n^2+1+3n)^2$
$= (n^2+3n+1)^2$ putting in standard form
This shall require that $a+1$ and $a+2$ do not increase the LCM.
This shall require both $a+1$ and $a +2 $to be composite and neither is a perfect power of a prime.
Let us explain it for $a+1$ and the same logic holds for $s+ 2$
If $a+1$ is prime then we have not encountered the same in any of the numbers and this shall contribute to LCM and it shall increase. LCM should be multiplied by $a+ 1$
If $a+1 = b^k$ where b is a prime then it should be multiplied by $b$ as $b^(k-1)$ must have come previously but not $b^k$
if $a+1=b^kc^m$ that is product of powers of 2 primes then $b^k$ and $c^m$ has already come and hence it shall not contribute to a higher LCM
Same for $a+2$
So we require smallest n such that n+1, n+ 2 are composite and producut of power of at least 2 primes
Looking at $1,2,3,4,5,6,7,8,9,10,11,12,13,14,15$
$14 = 2 * 7 $so LCM does not increase
$15 = 3 * 5$ so LCM does not increase
For $(8,9)$ both power of primes ,
For $(9,10)$, $9$ is power of prime
So $a = 13$
Next $a = 19$ as $20=2 ^2 * 5, 21 = 3 *7$ do not increase meets criteria
Without loss of generality we can assume that nmbers are in increasing sequence
We have 3 powers of 2 as 2 digit numbers . They are 16,32,64.
If we start at 10 we get 4 numbers in geometric progression when ratio is 2.
So we need some ratio less than 2 . Common ratio can be fraction as long as we do not get a fraction after multiplying by common ratio .
Because we need 5 numbers we need to multiply 4 times . So we start with a 4th power of 2 that is 16 and common ratio $\frac{3}{2}$ 16 giving 16, 24,36,54,81 and middle term is 36.
Basically 5 numbers $a^4,a^3b,a^2b^2,ab^3,b^4$ where a is starting number and $\frac{b}{a}$ as common ratio. form GP.
We shall prove a stronger result. we shall show that
$g(n) = n^{3k+2} + n^{3m+1} + 1\cdots(1)$ is not a prime for $n>1$ and $k+m > 0$
To prove it we shall snow that is is divisible by $n^2+n+1$
We have
$n^2+n+1 = (n-\omega)(n-\omega^2)..\cdots$ where $\omega$ is cube root of 1
And $w^3 = 1\cdots(3)$
And $w^2+w+1=0\cdots(4)$
We shall snow that g(n) is divisible by $ (n-\omega)$ and $ (n-\omega^2)$
We get putting $\omega$ in (1) for n
$g(\omega) = \omega^{3k+2} + \omega^{3m+1} +1$
$= (\omega^3)^k \omega^2 + (\omega^3)^m \omega + 1$
$= \omega^2 + \omega + 1 = 0$ using (3) and (4)
So $g(n)$ is divisible by $(n-\omega)$
Similarly $g(n)$ is divisible by $(n-\omega^2)$
So $g(n)$ is divisible by $n^2+n+1$
Because $k+m$ is greater than 0 so at least one of them is greater than zero
So $g(n) > n^2+n+1$ and as $n^2+n+1 > 0$ g(n) is product of 2 numbers neither is 1
Hence g(n) is composite
Putting k = m = 1 we get $f(n)$
So f(n) is not prime
Proved
Let two integers be $a^2+ab+b^2$ and $c^2+cd+d^2$
We have
$a^2+ab+b^2 = (a-b\omega)(a-b\omega^2)\cdots(1)$ where $\omega$ is complex cube root of 1
And
$c^2+cd+b^2 = (c-d\omega)(c-d\omega^2)\cdots(2)$
Multiplying (1) by (2) we get
$(a^2+ab+b^2)( c^2+cd+d^2) = (a-b\omega)(a-b\omega^2)(c-d\omega)(c-d\omega^2)$
Now as $\omega$ is cube root of 1 so we have
$\omega^2+\omega+1 = 0$
Or
$\omega^2 = - (1+\omega)\cdot(3)$
Also
$\omega = - (1+\omega^2)\cdot(4)$
Now get us calculate $(a-b\omega)(c-d\omega)$
$(a-b\omega)(c-d\omega) = ac -\omega(ad + bc) + bd\omega^2$
$= ac -\omega(ad+bc) -bd(1+\omega)$ from (3)
$= (ac-bd) -\omega(ad+bc+bd)\cdots(5)$
Now get us calculate $(a-b\omega^2)(c-d\omega^2)$
$(a-b\omega^2)(c-d\omega^2) = ac -\omega^2(ad + bc) + bd\omega^4$
$= ac -\omega^2(ad+bc) +bd(\omega)$ as $\omega^3=1$
$= (ac -\omega(ad+bc) +bd(1+\omega^2)$ from (4)
$= (ac-bd) -\omega^2(ad+bc+bd)\cdots(6)$
Now $(a^2+ab+b^2)( c^2+cd+d^2) = (a-b\omega)(a-b\omega^2)(c-d\omega)(c-d\omega^2)$
$=(a-b\omega)(c-d\omega)(a-b\omega^2)(c-d\omega^2)$
$=((ac-bd) -\omega(ad+bc+bd))((ac-bd) -\omega^2(ad+bc+bd))$
$=((ac-bd)^3 +(ac-bd)(ad+bc+db)+(ad+bc+db)^3$ using $x^3+y^3=(x-\omega y)(x-\omega^2 y)$
= $m^2+mn + n^2$ where $m= ac-bd$ and $n = ad+bc+bd$
Hence proved
We are given
$6a+10b+15c=3000\cdots(1)$
We have rearranging the terms let us get a in terms of others
$6a = 3000 - 10b-15c = 5(600-2b-2c)$
As the RHS is divisible by 5 and GCD(5,6) = 1 so a is divisible by 5 and so
$a=5x\cdots(2)$ for some x
Again to get b in terms of others
$10b = 3000 - 6a-15c = 3(1000-2a-5c)$
As the RHS is divisible by 3 and GCD(3,10) = 1 so b is divisible by 3 and so
$b=3y\cdots(3)$ for some y
Again to get c
$15c = 3000 - 6a-10b = 2(1500-3a-5b)$
As the RHS is divisible by 2 and GCD(2,15) = 1 so c is divisible by 2 and so
$c=3z\cdots(4)$ for some z
Putting the values a,b,c from (2),(3),(4) respectively in (1)
$30x + 30y + 30z = 3000$
Or $x + y + z = 100$
We need to find x,y,z all 3 positive integers sum is 100
Let us assume that there are 100 stones in a line . There are 99 gaps. we can make them into 3 parts of each part containing above one by putting 2 sticks in 99 gaps. this can be done in $99 \choose 2$ ways
Before finding the n let us make a few obsevations
We have for any n $9 | 10^n-1$
That is $3^2 | 10^n-1$
And as $ 3| 111$ or $3 | \frac{10^3-1}{9}$
Hence $3^3 | 10^3-1$
Now we need to move from $3^3$ to $3^5$
999 is divisible by 27
Appending another 999 to the left shall give 999999 and it is divisible by 27 appending 9 or 99 does not make if divisible by 27
But appending 999 to the left is adding $999 * 10^3$
As $243 = 27 *9$ we need to keep a-dding until the
Let us keep appending say n terms
We get $999(1+ 10^3 + 10^6 + \cdots 10^{3n})$
For this to be divisible by 243 as 999 is divisible by 27 $1+ 10^3 + 10^6 + \cdots 10^{3n})$ must be divisible by 9.
Each term of $1+ 10^3 + 10^6 + \cdots 10^{3n})$ divided by 9 leaves a remainder 1 so there should by 8 terms and so n= 24
Then we get the number = $999(1+10^3 + \cdots 10^24)$
$= (10^3-1)(1+10^3 + \cdots 10^24)= 10^27-1$
so n= 27
We have
$1280000401 = 1280000000 + 400 + 1= 20^7 + 20^2+ 1$
This is $x^7 + x^2 + 1$ where x = 20
All the terms have coefficient 1
There is an exponent 7 and and an exponent 2 one if of the form 3n + 2 and another of the form 3n+1
This is divisible by $x^2+x+1$ hence composite
Let us prove the same.
$x^7 + x^2 + 1$
$= x^7 -x + x^2 + x + 1$
$=x(x^6-1) + x^2 + x + 1$
$=x(x^3+1)(x^3-1) + (x^2 + x + 1)$
$=x (x^3+1)(x-1)(x^2+x+1)+ x^2 + x + 1$
$=(x^2+x+1) (x(x^3+1)(x-1) + 1$
We have factored the same and
so the given number is composite
Now Let us factor 1994
$1994 = 2 * 997$
We need to show that it is divisible by 2 and 997
RHS is even so is is divisible by 2
We need to show that the RHS is divisible by 997
We have as 997 is close to 1000 so let is brrin power of 10 and 2 to be as close as 997
$10^3 \equiv 3 \pmod {997}\cdots(1)$
and $2^{10} \equiv 27 \pmod {997}\cdots(2)$
so $10^{900} - 2^{1000}$
$=({10^9})^{100} - (2^{10})^{100}$ )
This is divisible by $10^9 - 2^{10}$
Now $10^9 - 2^{10}\pmod {997}$
$= (10^3)^3 - 2^{10}\pmod {997}$
$= 3^3 - 27 \pmod {997}$
$= 27-27 \pmod {997}$
= 0
So the number is divisible by 997
As it is divisible by 997 and 2 so it is divisible by 1994
The sequence is the squares of numbers which leaves a remainder 56 when divided by 100 that is
$n \equiv 56 \pmod {100}$
We can look at the numbers mod 100 and then find the numbers from table lookup. However this is not interesting
To keep the things simple let us find the numbers mod 10 and then make it to numbers mod 100.
Let us look at the number mod 10 and find the square mod 10 we get (1st is number mod 10, 2nd is number square mod 10 $(0,0),(1,1),(2,4),(3,9),(4,6),(5,5),(6,6),(7,9),(8, 4),(9,1)$
So then umber is 4 mod 10 or 6 mod 10
That is number is of the form 10n+ 4 or 10n+ 6
Now we need to find n in both the forms.
We can take the x as 2 digit number . adding 100 to x shall not change the 2 digits mod 100
This is because
$(100n + k)^2 = 10000n^2 + 200nk + k^2$
so $(100n + k)^2 = k^2\pmod {100}$
Consider first form $x = 10n + 4$
Now $(10n+4)^2 = 100n^2 + 80n + 16$
so $80n + 16 = 56\pmod {100}$
or $80n = 40 \mod 100$
dividing by 20 on both sides
$2n = 1\pmod 5$
or $n = 3 \mod 5$ or $n \in \{3,8\}$ as we are looking for single digit number
So the number is $34 \pmod {100}$ or $84 \pmod {100}$
we can combine both to get $x \equiv 34\pmod {50}$
Considerfnext form $x = 10n + 6$
Now $(10n+6)^2 = 100n^2 + 120n + 36$
As remainder is 56
so $20n + 16 = 56\pmod {100}$
or $20n = 20 \mod 100$
dividing by 20 on moth sided
$n = 1\pmod 5$
or $n = 1 \mod 5$ or $n \in \{1,6\}$ as we are looking for single digit number
So the number is $16 \pmod {100}$ or $66 \pmod {100}$
So we have $x \equiv 16 \pmod {50}$ or $x \equiv 34 \pmod {50}$
As RHS is power of 2 so both $x-8$ and $ x-10$ should be power of 2 or they should be -ve of power or 2 because there are two terms on the left hand side so product shall be positive.
As $(x-8) - (x-10) = 2$ so the 2 values should be 4 and 2 or -2 and 4( because 2 and 4 the power of 2 which differ by 2)
Naturally x-8 shall be lower value
So taking $x -8 = 4$ or $x = 12$ we get $2^y= 4* 2 = 8$ or $y = 3$ giving $((x,y) = (12,3)$
Taking $x-8 = -2$ we get $x = 6$ and $y = 3$ or $((x,y) = (6,3)$
So we have 2 set of solutions $(x,y) \in \{(12,3),(6,3)\}$If is is divisible by 12 then it is divisible by 3 and also by 4. this is so because 3 and 4 are co-primes
First Let us prove that is divisible by 3
Let us take $(b-a) , (c-a), (d-a)$ mod 3
As taking mod 3 there are 3 different remainders there are 2 possibilities
Having proved divisible by 3 Now let us prove that is is divisible by 4
Let us take $(b-a) , (c-a), (d-a)$
If at least two of them are even then it is divisible by 4
1. If one is even say $d-a$ then $b-a$ and $c-a$ are odd and $(c-a) - (b-a)$ that is c-a is even . As there are 2 even numbers so product is divisible by 4
2. If none is even the $b-a$ and$ c-a$ being odd $c-b$ is even and $b-a$ and $d-a $being odd $d-b$ is even . As $c-b$ and $d-b$ are even so product is divisible by 4
In all cases it is divisible by 4
Hence it is divisible by 12
We need to show that $(a^2+b^2+ c^2)^2$ can be expressed as sum of 3 squares.
Let for simplicity
$x= a^2+b^2\cdots(1)$
and
$y=c^2\cdots(2)$
So we have
$(a^2+b^2 + c^2)^2 = (x+y)^2$ from (1) and (2)
$=(x-y)^2 + 4xy$ as $(x+y)^2 - (x-y)^2 = 4xy$
$= (a^2+b^2-c^2)^2 + 4(a^2+b^2)c^2$ putting back values
$= (a^2+b^2-c^2)^2 + 4a^2c^+4b^2c^2$
$= (a^2+b^2-c^2)^2 + (2ac)^2 + (2bc)^2$
Hence proved
We shall prove it by Principle of Mathematical Induction
The base step is n = 3 and n = 4
Because it is right angled triangle and c is the hypotenuse we have
$a^2 + b^2 = c^2\cdots(1)$
Further as
$a < c$
Multiplying both sides by a^2 we have
$a^3 < ca^2\cdots(2)$
Similarly
$b^3 < cb^2\cdots(3)$
Adding (2) and (3) we get
$a^3 + b^3 < ca^2 + cb^2$
or
$a^3+b^3 < c(a^2+b^2)$
or $a^3 + b^3 <c^3$ using (2)
Additionally
Squaring (1) we get
$a^4 + b^4 + 2ab = c^4$
And hence dropping 2ab(which is positive) from the left
$a^4 + b^4 < c^4$
So we have proved the base step
For induction step we shall not move from n to n+1 but ftom n to n+ 2
We have Let t be true upto k = n
from k = n- 1 (not k) we shall prove for n+ 1
we have
$a^{k-1} < c^{k-1}$
Multiplying by a^2 on both sides we get
$a^{k +1} < c^{k-1}a^2\cdots(4)$
Similarly
$b^{k +1} < c^{k-1}b^2\cdots(5)$
from (4) and (5) we get
$a^{k +1} + b^{k+1} < c^{k-1}a^2 + c^{k-1}b^2$
Or
$a^{k +1} + b^{k+1} < c^{k-1}(a^2 + b^2)$
or $a^{k +1} + b^{k+1} < c^{k-1}c^2)$
or $a^{k +1} + b^{k+1} < c^{k+1}$
So the induction step is complete
We have proved for 3 and 4 ad from n-1 to n+ 1 and hence for all above 2
As we have proved base step and induction step proof is complete
$
Before we look for solutions let us quickly check based on modular arithmetic.
Looking at mod 3 we have 2022 is 2 mod 4.
Solution may exist.
Let us check based on mod 3.
We have
$2022 \equiv 0 \pmod 3$
But $2022 \equiv 6 \pmod 9$
So 2022 is not sum of 2 perfects squares
Let us prove the basis of the same
x is of the form $3a$ or $3a+1 or $3a+2$
If $x \equiv 0 \pmod 3$ then $x^2 \equiv 0 \pmod 3$
If $x \equiv 1 \pmod 3$ then $x^2 \equiv 1 \pmod 3$
If $x \equiv 2 \pmod 3$ then $x^2 \equiv 1 \pmod 3$
Similarly for y
Now looking at above we have $x^2+y^2 \equiv 0 \pmod 3 $ iff $x \equiv 0 \pmod 3$ and $y \equiv 0 \pmod 3$.
So x = 3a and y = 3b for some a and b
or $x^2+y^2 = 9(a^2+b^2)$ or divisible by 9
But 2022 is not divisible by 9
Hence no solution
Because LCM of 20 and 16 is 80. we need to look upto 80 an then the pattern repeats
Now remainder shall keep increasing till we get next multiple . then the remainder resets
Because we need to compare the remainder the number is of the form 20a + b and 16c + d ( 0 <=b <=19) , (o <=d <=16)
The remainder divided by 20 is lower than remainder divided by 16 if multiple of 16 is lower than multiple of 20. that is $16c < 20a < n < 16(c+1) $, That is the number lies between multiple of 20 and multiple of 16. The numbers that fit the condition are from 20a to 16(c+1)
Let us write multiples of 16 : $16,32,48,64,80$
multiple of 20 : -$20,40,60,80$
So the ranges are $0, 16(16*1),20(20*1),32(16*2),40(20*2),48(16*3),60(20*3),64(16*4),80$
the ranges where the remainder divided by 20 is lower in the rage $[20*1,16*2), [20*2,16 * 3]$ and $[20*3,16 * 4)$ in each range lower number is inclusive and higher is not giving $16*2 - 20 *1 + 16*3 - 20*2 + 16*4 - 20 *3$ or $24$ numbers.
In each lock of 80 numbers there are 24 numbers
upto 2000 there are 25 blocks so 25 * 24 600 numbers
From 2001 to 16 there are 16 number having same remainder as divided by 16 as divided 20
So no extra
So Ans 600
To factor this we shall take it as this as a polynomial of Let us write in decreasing power of x
$2x^2−y^2−3z^2−xy+4yz+5zx$
$=2x^2+x(-y+5z) -(y^2–4yz+3z^2)$
$= 2x^2+x(-y+5z) -(y-z)(y-3z)$ by factoring the the term independent of z.
We let $a = y-z$ and $b= y-3z$ which are factors of the independent term and they are co prime to each other
Now a and b are co-primes so 2 the coefficient of $x^2$ $must go with one of them and we have to chose the same so that the sum (or difference) is co-efficient of x.
We have $a - 2b = -y + 5z$
So we get
$2x^2 +(a-2b) x - ab$
$= 2x^2 + ax - 2bx -ab$+
$= x(2x+a)-b(2x+ a)$
$= (x-b)(2x+a) = (x-y+3z)(2x+y-z)$
It is product of 100 terms.
From each term we shall have 2 solutions.
Either both the solutions are real or complex
Let check which and how many terms shall have real roots
$x^2 - 11x +k$ has real root if discriminant is positive or zero
That is $11^2 - 4 * k \ge 0$ or $ k \le 30$
There are 30 terms for which it has got real roots and sum of real roots in each term is 30 giving sum of real roots 330
