₦airaland Forum

jackpot: You said infinitely more? Can you prove it?

;
yes

Alpha Maximus: Is there any correlation between such a triple and its successive/previous triple? Is there a general formula for obtaining such triples? Is there any correlation between the first, second and third terms of consecutive triples of this sort?
For example I discovered an easy way of producing Pythagorean Triples sometime ago, the formula is as follows and is sure-fire : 3n, 4n, 5n....where n is ANY number at all! It always works , try it! Assuming n=10, we'll have a Pythagorean triple of 30,40 and 50. Is this valid? Let's confirm.....30²+40²=50²
900+1600=2500
2500=2500 ....thus , the validity has been authenticated . 'N' can even be a decimal to 13 decimal places and it would still work. Laplacian, do you have any formula of this sort to obtain your cited triples in which the sum of one and the product of any two numbers in such a triple will result in the yield of a perfect square?

yes!

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!

what does "NL" mean? Numerical Language?

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......

...integers which satisfy my question in increasing order of the magnitude of their first term include;
1, 8, 15
4, 30, 56
15, 112, 127
56, 418, 780
2911, 21728, 40545
...infinitely more...

Much thanks 2 jackpot, 2Dx, 2nioshine & and all others who secretly wrkd on my 3yrs-old question, 4 ur gr8 concern

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, 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?...
Here's the full solution...with a=y-x d=y+x, z=2y+x slotted into either eqn1, eqn2 or eqn3, we arrive arrive @ the followin eqn
2y²-2xy+1=x²
or rearrangin 2 get
3y²+1=x²+2xy+y²
or
3y²+1=(x+y)²
but d=x+y
so
3y²+1=d²
or
d²-3y²=1
remember that since a=y-x
==> a=y+x-2x=d-2y...every other thing follows from the Pell's Equation above...Regards 2 jackpot and 2Dx...

...xsup2/sup

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

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}

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

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

jackpot: 1, 8, 15.

honestly, u dont knw hw gratful i am...u wont beliv the lenght of tim i seachd 4 a solution....Thanks....hw did u get them?

jackpot: 1, 8, 15.

honestly, u dont knw hw gratful i am...u wont beliv the lenght of tim i seachd 4 a solution....Thanks....hw did u get them?

smurfy: Beautifully correct!

Actually, it's such a simple question. The answer is 5!/2 = 60.

How? Well, let's look at the letters A, B, C. Total possible arrangements are ABC, ACB, BAC, BCA, CAB, CBA. Total possible arrangements is 6. In how many ways, for example, is C to the right of A? Three ways.

You can pick any two letters and you'll always get n!/2 ways of having a particular letter to the right/left of another. This is always so since n! is always even for any positive integral value of n greater than 1.

A big THANK YOU to akpos4uall, Laplacian and aysuccess99 for doing justice to this question.

what u have reasoned is correct because the arrangement is symmetrical....it follows immediately that we have deceitfully proved that;

{summation i=1 to n-1}
(n-2)P(i-1)*(n-i)!=n!/2

smurfy: In how many different ways can the letters P, Q, R, S, T be arranged if the letter R must be to the right of the letter S?

**Goes out to buy gift for whoever thrashes the question**

here is the general solution to your problem.....suppose u have r object and u 're required 2 find d number of ways of arrangin them so that one is always 2 d right of another....dig 'r' hole, from ur right, label any hole of ur choice as the i-th hole which is d hole occupied by d object required 2 b on d right...now, from hole 1 to i-1 only r-2 objects can fil it, (because an object is already occupyin d
i-th and d other object is 2 its left) so number of ways of arrangemnt of objects to the right of i is (r-2)P(i-1)....to the left of the i-th hole, only (r-i) holes are left for the same number of objects, so number of ways of arrangin objects to d left of i is (r-i)!
So totaly, number of ways of permutation,
n={summation i=1 to r-1}
(r-2)P(i-1)*(r-i)!

@Smurfy, for ur case above r=5, so we hav
n=3P0*4!+
3P1*3!+
3P2*2!+
3P3*1!
Or
n=24+18+12+6=60

try it 4 any value of r, u 'll obtain the correct answer

jackpot: Hi Mr Laps, the numbers are 0, 2, 4.

thanks ma'am...but only non-zero integers are allowed

[quote author=smurfy][/quote]case1: if R appears as first letter to d right,
number of ways of arranging d other letters, n=4!=24
case2: if R appears as second letter to d right,
n=4P3*3P1=24*3=72
case3: if R appears as third letter to d right,
n=4P2*3P2=12*6=72
case4: if R appears as fourth letter to d right,
n=4P1*3P3=4*6=24

so there are 192 possible ways of arrangement

Generally, if u are to arrange r objects so that one is always to d right of the other, then
case i: if object appears as i-th object to d right,
n=(r-1)P(r-i)*(r-2)P(i-1)

smurfy: Are you sure the question isn't talking about any two consecutive terms?

no

benbuks: ..hint

we can have a , (a+d) ,( a+2d)

remember to have a perfect square b^2 -4ac=0 (from theorem of quadratic equation) , hope it helps ,?
we have any series of ood/even numbers
e.g 3,5,7,. . .

tanx but i dont no hw 2 apply it here;
3*7+1=22

smurfy: Is it 2, 4, 6?

(2 * 4) + 1 = 9
(4 * 6) + 1 =25

Remember also that if a, b, c are in AP, then b = 1/2(a + c).

but 2*6+1=13

...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......

jauntee02: i'l give only the first person to solve this (within 12 hours) a recharge card.

1..The current distance between
Buenos Aires and Cape Town at
latitude 34° S is 6890 km, and
the sea floor spreading rate at
the South Atlantic mid-ocean
ridge (i.e. between the African
and South American plate) is 35
mm yr-1. Assuming that both
cities lie near to the edge of their
respective continental margins,
calculate to the nearest million
years how long it has been since
these two continents separated.
*****Hint: it may help if you first
calculate how far the plates
would move in 1 million years.
2..A palaeontologist has
discovered a new fossil
Pterosaur species (i.e. flying
reptile) in sedimentary strata
near Buenos Aires, and also near
Cape Town. The strata are dated
as Early Cretaceous (120 Ma) in
age. Using a reconstruction of
the fossil skeleton, the
palaeontologist has suggested
that the animal probably ate fish,
and could have flown long
distances across the sea at an
average of 60 km h-1. To the
nearest hour, determine how
long, it would have taken the
Pterosaur to fly across the Early
Cretaceous Atlantic Ocean.

1.)...the seperation rate,
r=35mm yr-1=35*10^-3 m yr-1
or
r=35*10^-3*10^6/10^6 m yr-1
or
r=35 km/million-yrs....
So,
the time T, taken for a seperation of 6890 km is;
T=6890/35
or
T=196.857 million-yrs

2.) Note that 120 Ma means 120 Mega-annum...meaning
120 Million-yrs...that been being the yr of flight, it means the seperation distance S, in that period is S=35*120=4200 km...so the duration P, of the specie's flight is P=4200/60=70 hrs or equivalently it 'll take 2-days 22-hours

you can keep ur Recharge Card

benbuks: X^4 + y^4 =82 .....(a)
x+y = 4 .....(b)
solve.

x^4+y^4=82 ....eqn1
and
x+y=4 ......eqn2
rearrang 2 get
x^4-1=81-y^4....eqn3
and
x-1=3-y........eqn4
or
(x-1)(x+1)(x^2+1)=(3-y)(3+y)(9+y^2)......eqn3
and
x-1=3-y........eqn4
or
(5-y)(17-8y+y^2)=(3+y)(9+y^2)
or
85-57y+13y^2-y^3=27+3y^2+9y+y^3
or
2y^3-10y^2+66y-58=0
or
y^3-5y^2+33y-29=0
the only real value of y is y=1
and x=3
not 4 ur prize though

benbuks: answer =(1 ,1 ,1) ...from (1) square both sides.
We have (x+y+z)^2 =9
but, (x+y+z)^2 =x^2 + y^2 +z^2 + 2(xy+ yz+ zx)

==>9=3 + 2(xy+ yz + zx)
==> xy+ yz + zx=3 .....(4)
we should know that,
x^3 + y^3 + z^3 -3xyz=(x+y+z)(x^2 + y^2 + z^2 -xy-yz-zx)
==>3-3xyz=(3)(3-3)
which produce xyz=1.....(5)

now, let's have
x+y+z=3 . . . . .(1)
xy+yz+zx=3 . . .(4)
xyz=1 . . . .(5)

now, from (5)
yz=1/x and y+z=3-x (from (1)) pluck these in (4), we have 1/x + x(3-x) =3
...=> x^3 + -3x^2 +3x -1=0
by factor theorem , x=1trice
put these in (1) & (2)
we have
y+z=2 . . . .(6)
y^2 +z^2= 2 . . . .(7)
by substitution method
we have
y=1twice
==>z=1 (since z=2-1)

hence (1 , 1 , 1)

....Outstanding....

smurfy: Again, you are wrong. The correct answer is 2(x + 7)(x + 2)(5x + 1)(2x - 3).

Now, what did you do wrong?

he's correct...both results are equivalent

benbuks: ..wow thanks man, you just safed my job.

....*madam jackpot , a'v cracked it ma..,oya give me my allowance, else i 'll go on strike.*.....lolz.

y re u alwaz swappin ur gender?

kwakayekaa: Let Q>0 be the set of positive rational numbers. Let f : Q>0 ! R be a function satisfying
the following three conditions:
(i) for all x; y 2 Q>0, we have f(x)f(y)  f(xy);
(ii) for all x; y 2 Q>0, we have f(x + y)  f(x) + f(y);
(iii) there exists a rational number a > 1 such that f(a) = a.
Prove that f(x) = x for all x 2 Q>0.

hey, wat symbol is between f(x)f(y) and f(xy)?
Wats d symbol between f(x + y) and f(x) + f(y)?
Write this in words f : Q>0 ! R

jackpot: @Laplacian and Humphrey,

I said criticize, not to type question mark and to reply "GOOD".

Well, if you can draw pictorially two open sets(remember to use dotted lines for the edges), then you will understand this proof.

view d questn mark again:*

jackpot: I will provide a lazy solution to the problem sha.

A set is open if none of its points are boundary points.

Now, let P and Q be open sets. Then boundary(P) lies outside P and boundary(Q) lies outside Q.

Now, P n Q is a subset of P. P n Q is also a subset of Q. This means that boundary(P) and boundary(Q) lies outside P n Q.

BOUNDARY(P n Q) IS A SUBSET OF [BOUNDARY(P) u BOUNDARY(Q)]. Infact if x € boundary(P n Q), then either x € P or x € Q (BUT NOT BOTH AT THE SAME TIME) since P and Q are open sets.
Thus, boundary(PnQ) lies outside PnQ.

Hence, PnQ is open.


Feel free to critize this solution, folks.

?

BOUNDARY(P n Q) IS A SUBSET OF [BOUNDARY(P) u BOUNDARY(Q)]....u didnt back up d above assertion in ur proof and its a vital part of d proof...a more logical consequence of ur explanation should b;
BOUNDARY(P n Q) IS A SUBSET OF [(P) u (Q)]...bcuz since boundary(P) and boundary(Q) lies outside P n Q, it follows directly dat boundary PnQ lies in P u Q

again; (BUT NOT BOTH AT THE SAME TIME)...OBVIOUSLY does not follow from d fact dat P & Q are open: as an illustration, supposin Q lies wholly in P, then x€ boundary of both P and Q....which contradicts ur assertion...

jackpot: ^you are assuming that all your sets are subsets of IR, the set of real numbers and you are also assuming that elements can be compared by your use of less than sign. ( < )

my question is, what if the sets are open in R²? Or what if they are domains in the Complex Plane?

Can your proof work for an arbitrary open set?

@Humphry, i suggest u & Benbuks should always provide d incorrect steps in any wrong solutions like jackpot alwaz does and myselt too, not 2 simply dismis it & leave d solver in doubt&confusion...even if my solution does not cover d general case, it has @ least settled d case pointed out by jackpot abov....and dat should worth some apprreciation...
u 'r modest lady jackpot & i like ur styl of criticism....i 've not really wrked on provin d general case though, but i 'll giv it a second thought now that u hav underlined it...but i know it will hav 2 incoperate d NORM and follow a similar chain of reasonin as abov....by d way, wat's preventin u from supplying a proof? Or 're u just watchin 4rm d sidelines just 2 handpick my errors?....
@ALL, am not here 2 enter into competition wit anybdy, 4 those dat copy questions (witout havin challenges wit them) from textbooks & paste here.....i'll only attempt questions only by individuals who hav difficulties wit them & are genuinely desperate 4 their solution...

jackpot: Hi Sir Laplacian

how can a set be less than or equal to a point x?

sorry, i was tryin 2 avoid lengthy computations, i tot everybdy 'll undstnd.
put mathematically,
the set
P={p : x<=p<=y},

M={m : a<m<A},

N={n : b<n<B}.

Z={ z: y<z<A},

NB: z=(y+A)/2€Z
Every other thing remains the same

Humphrey77: sin2x+sinx is equal to 1 we know that sin2x is equal to 2sinxcosx. clearly; we have 2sinxcosx +sinx equal to 1 . so let sinx equal to y; by subtitution into

sin2x + sinx equal to 1 we obtain the equation: 4y"4-3y"2-2y+1 equal to zero ; by solving for y we obtain y equal to 1; clearly recall that sinx is equal to y so we have x eq
ual to arcsin(1) we have x equal to 90 so x is equal to 90

solvin d differential eqn u cannot get y=1,
suppose y=1, indeed y'=0, y"=0,
substitute into ur differential eqn,
4y"4-3y"2-2y+1=0 to obtain

0-0-2+1=0
or
-1=0

hw did u solve ur differential eqn?