akpos4uall:
continuing from here

2xy = 2z² - 6z + 6
:. xy = z² - 3z + 3

Having known the expression for xy in terms of z as written above. We can now substitute both expression (i.e the xpression for (x + y) and xy) in terms of z in equation 6. x + y = (3 - z) and xy = z² - 3z + 3

(x + y)³ - 3xy (x + y) + z³ = 3
substituting yields :
(3 - z)³ - 3( z² - 3z + 3) (3 - z) + z³ = 3

27 - 27z + 9z² - z³ + 3z³ - 18z²
+ 36z - 27 + z³ = 3

3z³ - 9z² + 9z - 3 = 0
Simplifying further:

z³ - 3z² + 3z - 1 = 0
Factoring the polynomial yields:
(z - 1)³ = 0
z = 1 thrice

lantessy: recheck d expression 4 ur xy and also z cannot be root 2...but u did a good job

Laplacian: If
a^2+ad+1=x^2.....eqn1

a^2+2ad+1=y^2......eqn2

a^2+3ad+2d^2+1=z^2.......eqn3

1.) find a in terms of x, y, and z

2.) find d in terms of x, y, and z

doubleDx:


a² + ad + 1 = x².....eqn1

a² + 2ad + 1 = y²......eqn2

a² + 3ad + 2d² +1 = z².......eqn3

1.) find a in terms of x, y, and z

2.) find d in terms of x, y, and z


I think this is the interpretation to the problem you posted earlier.... I arrived at something similar when I attempted solving the problem in the afternoon, too many unknowns !

@jackpot, how you take crack am? Trial and error method?....


1. Solution =>

From (1) a² = x² - ad - 1 .....(A)

Substituting (A) in (2) yields =>

ad = y² - x²............3

Substituting (A) in 3 yields =>

2ad + 2da² = z² - x²

Factoring the LHS =>

2ad( 1 + a) = z² - x²......4

Substituting (3) in (4) yields =>

2(y² - x²)(a + 1) = z² - x²

(a + 1) = (z² - x²)/2(y² - x²)

:. a = (z² - x²)/(2y² - 2x²) - 1

a = {(z² - x² - (2y² - 2x²)}/(2y² - 2x²)
a = {(z² - x² - 2y² + 2x²}/(2y² - 2x²)
a = {(z² + x² - 2y²}/(2y² - 2x²)

. Solution =>

From eqn (3) above

d = y² - x²/a

Substituting a = {(z² + x² - 2y²}/(2y² - 2x²)

.: d = (y² - x² )*2(y² - x²)/{z² + x² - 2y²}
.: d = 2(y² - x² )²/{x² - 2y² + x²}

Laplacian: @mentor dx, how did u apply the index notation witout using d symbol ^

Laplacian:
i really appreciate ur effort...u 're my mentor indeed.....i 've been waitin on jackpot 4 her approach/response but it seems she's playin hide&seek....if only she knew what d solution meant....anyway, i 'll giv it one more week after which i'll abandon it with/without the solution...once again, thanks

smurfy:

I believe your answers to Quetions 2(a) and 2(b) are incorrect. The n!/2 method used in dealing with the previous PQRST question doesn't work here.

This one is harder because of the 2 P's and 3 Q's. The stated condition is that the 2 P's must always be to the right of R.

For example, QSRTQPQP satisfies the stated condition in the question.

If the two letters under consideration were any of RST, then the 1/2[8!/(2!*3!)] idea would've worked.

=======
Here goes...

SOLUTION TO 2(A)

We need, always, a ...R....P...P... kind of arrangement.

Case 1: R Fills 1st Hole

The remaining seven holes can be filled in 7!/(2!*3!) = 420 ways.

Case 2: R Fills 2nd Hole

In how many ways can we fill the FIRST hole? There are two scenarios here: fill with Q or any of ST.

Filling the first hole with Q gives 6!/(2!*2!) = 180 ways.

Filling the first hole with ST (i.e. S or T) gives 2 * 6!/(2!*3!) = 120 ways.

So, the number of ways R fills the second hole is 180 + 120 = 300 ways.

Case 3: R Fills 3rd Hole

Now, we try to fill the FIRST TWO holes and then the last five holes to the right of R.

The four scenarios are: QQR*****, QSR*****, QTR*****, STR*****

QQR gives 1 * 5!/2! = 60
QSR gives 2 * 5!/(2!*2!) = 60
QTR gives 2 * 5!/(2!*2!) = 60
STR gives 2 * 5!/(2!*3!) = 40

So, the number of ways R fills the third hole is 60 + 60 + 60 + 40 = 220 ways.

Case 4: R Fills 4th Hole

[I think you now get the drift...]

The four scenarios are QQQR****, QQSR****, QQTR****, QSTR****

QQQR gives 1 * 4!/2! = 12
QQSR gives 3 * 4!/2! = 36
QQTR gives 3 * 4!/2! = 36
QSTR gives 6 * 4!/(2!*2!) = 36

So, the number of ways R fills the fourth hole is 12 + 3(36) = 120 ways.

Case 5: R Fills 5th Hole

The three scenarios are QQQSR***, QQQTR***, QQSTR***

QQQSR gives 4! * 3!/2! = 12
QQQTR gives 4! * 3!/2! = 12
QQSTR gives 6 * 3!/2! = 36.

So, the number of ways R fills the 5th hole is 12 + 12 + 36 = 60 ways.

Case 6: R Fills 6th Hole

Here, the arrangement is definitely of the form *****RPP. Therefore, we simply consider the number of ways of arranging the remaining letters QQQST to the LEFT of RPP. [Note that the 2 P's to the right of R can be arranged in 2!/2! = 1 way.]

This gives 5!/3! = 20 ways.

FINAL ANSWER

The number of ways of arranging the letters P, P, Q, Q, Q, R, S, T so that the letter R is (always) to the left of the letter P (i.e. the 2 P's) is 420 + 300 + 220 + 120 + 60 + 20 = 1140 ways.

=======

A special THANK YOU to akpos4uall for attempting Question 2.

doubleDx:


I made a little mistake, 2ad + d^2 = (z² - x²) AND not 2ad + da^2 = (z² - x²); so I'm re-posting =>

a² + ad + 1 = x².....eqn1

a² + 2ad + 1 = y²......eqn2

a² + 3ad + 2d² +1 = z².......eqn3

1.) find a in terms of x, y, and z

2.) find d in terms of x, y, and z


2. Solution =>

From (1) a² = x² - ad - 1 .....(A)

Substituting (A) in (2) yields =>

ad = y² - x²............3

Substituting (A) in 3 yields =>

2ad + 2d² = z² - x²

Factoring the LHS =>

2ad( a + d) = z² - x²......4

Substituting (3) in (4) yields =>

2ad + 2d² = z² - x²

2(ad + d²) = z² - x²

d² + ad = (z² - x²)/2

d² + y² - x² = (z² - x²)/2

d² = (z² - x²)/2 + (x² - y²)

d² = {(z² - x²) + 2(x² - y²)}/2

d² = {z² + x² - 2y²}/2

d = √[{z² + x² - 2y²}/2]


1. Solution =>

d = √[{z² + x² - 2y²}/2]

From (3) above

d = {y² - x²}/a............(5)

equating (5) and the value of d yields =>


{y² - x²}/a = √[{z² + x² - 2y²}/2]

:. a = {y² - x²}/√[{z² + x² - 2y²}/2]

a = [ {y² - x²}.√2{z² + x² - 2y²} ]/{z² + x² - 2y²}

jackpot: @Laplacian

abeg, na TaE method I used.

Anyways, I started the hunt by simplifying my choice for the first term, so I set a=1.

Next, I looked for the closest square: 4. I choose 3 as the second term so that 1(3)+1=4.
But the third term 5 didnt work since 1(5)+1=6, even though 3(5)+1=16.

So, I began shopping for the next closer square: 9. I then re-choose 8 as my second term so that 1(+1=9. Now, the third term is 15. 1(15)+1=16. 8(15)+1=121. Purrrrrrrrfect!

Took 2 mins though, but I know that my next search for the next triple may take hours. Maybe am lucky with these triples 1, 8, 15.

Simplifying my choice for a by setting "a=1" helped a lot since it tells you that the possible second term has to be "perfect square minus 1" for obvious reasons.

disclaimer: "a" can be any integer apart from 1 though. Using TaE method, it is good to fix the first term since the possible second terms comes handy. The third term is the "make or mar" term. For example in 1,3,5, the term 5 marred it since 1(5)+1 is not a perfect square.


Next, it is interesting to note that any three consecutive even/odd numbers(except 0,2,4) nearly satisfies it. This is an indirect consequence of difference of two squares. Maybe this "nearly" property triggered your research?

Proposition
No three consecutive even/odd numbers satisfies Laplacian's question

Proof
Let a be an integer and common difference d=2.
Then the three consecutive even/odd numbers are
a, a + 2, a +4

Now,
a(a+2)+1=a²+2a+1=(a+1)²,
which is a perfect square.
(a+2)(a+4)+1=a²+6a+9=(a+3)²,
again, a perfect square.
[s]two pefect squares already. No wonder the "nearly satisfication" claim[/s]

But,
a(a+4)+1=a²+4a+1=(a+2)²-3,
which is not, in general, a perfect square, except if (a+2)²=4, in which case a=0, a+2=2, a+4=4 (or a=-4, a+2=-2, a+4=0). No wonder 0, 2, 4 and -4, -2, 0 are nice brides

The proof is complete.










Lemme stop here.

doubleDx:


I made a little mistake, 2ad + d^2 = (z² - x²) AND not 2ad + da^2 = (z² - x²); so I'm re-posting =>

a² + ad + 1 = x².....eqn1

a² + 2ad + 1 = y²......eqn2

a² + 3ad + 2d² +1 = z².......eqn3

1.) find a in terms of x, y, and z

2.) find d in terms of x, y, and z


2. Solution =>

From (1) a² = x² - ad - 1 .....(A)

Substituting (A) in (2) yields =>

ad = y² - x²............3

Substituting (A) in 3 yields =>

2ad + 2d² = z² - x²

Factoring the LHS =>

2ad( a + d) = z² - x²......4

Substituting (3) in (4) yields =>

2ad + 2d² = z² - x²

2(ad + d²) = z² - x²

d² + ad = (z² - x²)/2

d² + y² - x² = (z² - x²)/2

d² = (z² - x²)/2 + (x² - y²)

d² = {(z² - x²) + 2(x² - y²)}/2

d² = {z² + x² - 2y²}/2

d = √[{z² + x² - 2y²}/2]


1. Solution =>

d = √[{z² + x² - 2y²}/2]

From (3) above

d = {y² - x²}/a............(5)

equating (5) and the value of d yields =>


{y² - x²}/a = √[{z² + x² - 2y²}/2]

:. a = {y² - x²}/√[{z² + x² - 2y²}/2]

a = [ {y² - x²}.√2{z² + x² - 2y²} ]/{z² + x² - 2y²}

Laplacian: ...xsup2/sup

Laplacian:
@jackpot, 2nioshine&Dx....tanks 4 sharin my burden...believin everyone understnds Dx's solutn, i proceed 2 give my solutn cuz i went farther & wait 4 ur correctns...
Now
d = √[{z² + x² - 2y²}/2]

but d is an integer, so the square root has to go...
Set z=2y+x, then
z²=4y²+4xy+x²
addin
x² - 2y² to both sides and dividin the result by 2 we obtain
[{z² + x² - 2y²}/2]=(x+y)²
hence
d=√{(x+y)²}

d=x+y

but, a=(y²-x²)/d
so that
a=(x-y)*(x+y)/(x+y)=x-y
so for integers x, y & z, the first term a=x-y, and the common differenc d=x+y and lastly z=2y+x

if nw we set x=3, y=4 then z=2*4+3=11, a=4-3=1, d=4+3=7
so the sequence is a=1,
a+d=1+7=8, a+2d=1+14=15

i.e 1, 8, 15.....which is jackpot's result....but i dont knw y it fails for other values of x and y...any suggestion plz?...

2nioshine: Exactly bruv am glad u made the correction urself...although we took a different route but got same end.....still in my Obs mood...nice work at all...stress to paste solution no be childs'

smurfy:

I believe your answers to Quetions 2(a) and 2(b) are incorrect. The n!/2 method used in dealing with the previous PQRST question doesn't work here.

This one is harder because of the 2 P's and 3 Q's. The stated condition is that the 2 P's must always be to the right of R.

For example, QSRTQPQP satisfies the stated condition in the question.

If the two letters under consideration were any of RST, then the 1/2[8!/(2!*3!)] idea would've worked.

=======
Here goes...

SOLUTION TO 2(A)

We need, always, a ...R....P...P... kind of arrangement.

Case 1: R Fills 1st Hole

The remaining seven holes can be filled in 7!/(2!*3!) = 420 ways.

Case 2: R Fills 2nd Hole

In how many ways can we fill the FIRST hole? There are two scenarios here: fill with Q or any of ST.

Filling the first hole with Q gives 6!/(2!*2!) = 180 ways.

Filling the first hole with ST (i.e. S or T) gives 2 * 6!/(2!*3!) = 120 ways.

So, the number of ways R fills the second hole is 180 + 120 = 300 ways.

Case 3: R Fills 3rd Hole

Now, we try to fill the FIRST TWO holes and then the last five holes to the right of R.

The four scenarios are: QQR*****, QSR*****, QTR*****, STR*****

QQR gives 1 * 5!/2! = 60
QSR gives 2 * 5!/(2!*2!) = 60
QTR gives 2 * 5!/(2!*2!) = 60
STR gives 2 * 5!/(2!*3!) = 40

So, the number of ways R fills the third hole is 60 + 60 + 60 + 40 = 220 ways.

Case 4: R Fills 4th Hole

[I think you now get the drift...]

The four scenarios are QQQR****, QQSR****, QQTR****, QSTR****

QQQR gives 1 * 4!/2! = 12
QQSR gives 3 * 4!/2! = 36
QQTR gives 3 * 4!/2! = 36
QSTR gives 6 * 4!/(2!*2!) = 36

So, the number of ways R fills the fourth hole is 12 + 3(36) = 120 ways.

Case 5: R Fills 5th Hole

The three scenarios are QQQSR***, QQQTR***, QQSTR***

QQQSR gives 4! * 3!/2! = 12
QQQTR gives 4! * 3!/2! = 12
QQSTR gives 12 * 3!/2! = 36.

So, the number of ways R fills the 5th hole is 12 + 12 + 36 = 60 ways.

Case 6: R Fills 6th Hole

Here, the arrangement is definitely of the form *****RPP. Therefore, we simply consider the number of ways of arranging the remaining letters QQQST to the LEFT of RPP. [Note that the 2 P's to the right of R can be arranged in 2!/2! = 1 way.]

This gives 5!/3! = 20 ways.

FINAL ANSWER

The number of ways of arranging the letters P, P, Q, Q, Q, R, S, T so that the letter R is (always) to the left of the letter P (i.e. the 2 P's) is 420 + 300 + 220 + 120 + 60 + 20 = 1140 ways.

=======

A special THANK YOU to akpos4uall for attempting Question 2.

Laplacian: ...pls i need help with this questoin...either a solution or an idea is welcome...

Three non-zero integers are in arithmetic progression, the product of ANY two when when increased by one is a perfect square...find the numbers......

Alpha Maximus: Z²......finally got the hang of this NL index notation system! Ok lemme freestyle:
Standard Quadratic Equation Form: ax²+bx+c=0
Fermat's Last Conjecture states that no three positive integers a, b and c can satisfy the equation aⁿ+bⁿ=cⁿ for any positive integer values of n greater than 2.
Pythagoras' Theorem states that the sum of the squares of the smaller sides of a triangle is equal to the square of the hypotenuse. Mathematically, a²+b²=c²....Yes!!! Index notation overusage, here I come!

Laplacian:
what does "NL" mean? Numerical Language?

Alpha Maximus: Z²......finally got the hang of this NL index notation system! Ok lemme freestyle:
Standard Quadratic Equation Form: ax²+bx+c=0
Fermat's Last Conjecture states that no three positive integers a, b and c can satisfy the equation aⁿ+bⁿ=cⁿ for any positive integer values of n greater than 2.
Pythagoras' Theorem states that the sum of the squares of the smaller sides of a triangle is equal to the square of the hypotenuse. Mathematically, a²+b²=c²....Yes!!! Index notation overusage, here I come!

Alpha Maximus: Was about to ask the same question. But Numerical Language would have also been a good cover-up!