For those who may be confused, what he has written mayb replicatd as follows;

Alpha Maximus: For example, 19 is happy, as the associated sequence is:

1² + 9² = 82
8² + 2² = 68
6² + 8² = 100
1² + 0² + 0² = 1.
The 143 happy numbers up to 1,000 are:

1, 7, 10, 13, 19, 23, 28, 31, 32, 44, 49, 68, 70, 79, 82, 86, 91, 94, 97, 100, 103, 109, 129, 130, 133, 139, 167, 176, 188, 190, 192, 193, 203, 208, 219, 226, 230, 236, 239, 262, 263, 280, 291, 293, 301, 302, 310, 313, 319, 320, 326, 329, 331, 338, 356, 362, 365, 367, 368, 376, 379, 383, 386, 391, 392, 397, 404, 409, 440, 446, 464, 469, 478, 487, 490, 496, 536, 556, 563, 565, 566, 608, 617, 622, 623, 632, 635, 637, 638, 644, 649, 653, 655, 656, 665, 671, 673, 680, 683, 694, 700, 709, 716, 736, 739, 748, 761, 763, 784, 790, 793, 802, 806, 818, 820, 833, 836, 847, 860, 863, 874, 881, 888, 899, 901, 904, 907, 910, 912, 913, 921, 923, 931, 932, 937, 940, 946, 964, 970, 973, 989, 998, 1000
Culled From Wikipedia

My little observations, proofs to follow shortly;

1.) if a number is HAPPY, then, that number multiplied by any integral power of ten is also HAPPY. e.g 7 and 70 and 700
2.) if a number is HAPPY, the number formed by permutation of the digits of the original number is also HAPPY. e.g 19 and 91 or 49 and 94
3.) the product of any two HAPPY numbers may/may not be HAPPY. e.g 7*19=133
4.) a binary number is HAPPY iff the nimber of 1 it contains is itself HAPPY
5.) the set of ALL HAPPY numbers is not closed under the multiplicative binary operation, and so is not a Group.
6.) the sum and difference of any two HAPPY number is not a HAPPY number
7.) if two numbers are HAPPY, then the number formed by combinin the digits of the two original numbers is not HAPPY

Humphrey77: WHAT ARE ALGEBRAIC NUMBERS? :*@HAPPY

Algebraic numbers is a number that is a root of a non zero polynomial in one variable with rational coefficients or equivalent by clearing denominators with integer coefficients.

rhydex 247:

Algebraic numbers is a number that is a root of a non zero polynomial in one variable with rational coefficients or equivalent by clearing denominators with integer coefficients.

Illustrative examples please!

solve for h in the equation : 6(raise to power h) -31(2 raise to power h ) +32 is equal to zero.

pls sum1 shuld asist in solvin d integral of exponential of x to d power of two

Laplacian: For those who may be confused, what he has written mayb replicatd as follows;



My little observations, proofs to follow shortly;

1.) if a number is HAPPY, then, that number multiplied by any integral power of ten is also HAPPY. e.g 7 and 70 and 700
2.) if a number is HAPPY, the number formed by permutation of the digits of the original number is also HAPPY. e.g 19 and 91 or 49 and 94
3.) the product of any two HAPPY numbers may/may not be HAPPY. e.g 7*19=133
4.) a binary number is HAPPY iff the number of 1 it contains is itself HAPPY
5.) the set of ALL HAPPY numbers is not closed under the multiplicative binary operation, and so is not a Group.
6.) the sum and difference of any two HAPPY number is not a HAPPY number
7.) if two numbers are HAPPY, then the number formed by combinin the digits of the two original numbers is not HAPPY
.

proofs:

PROPOSITION 1:

proof: let xy be a two digit HAPPY number, then the followin sequenc holds;
x²+y²=abcd

a²+b²+c²+d²=mnpqsr
.....
.....
..... =10000....0

Suppose we have xy*10^k for some integer k, then
xy*10^k=xy00000...0 with the zeros in k places, hence
x²+y²+0²+0²+...+0²=abcd, and the above sequenc is repeatd and the result follows.

COROLLARY 1: there are infinitely many HAPPY numbers.

Proof; because k is a natural number and there are infinitely many natural numbers, the result follows

PROPOSITION 2:

proof: let xyz be HAPPY, then
x²+y²+z²=abcd, where abcd is HAPPY, now the permutation yxz is shows that y²+x²+z²=x²+y²+z²=abcd, {because addition is cummutative}, and the result follows..

COROLLARY 2: if any number of zeros is either struck out or fixed in between the digits of a HAPPY number, the new number formed is also HAPPY.

Proof:
case 1: let the number be
xyz0000...0 havin k zeros, then by the above proposition, the number can be permuted to take m zeros to the left of the natural numbers, i.e
000...0xyz000..0=xyz000...0, where the zeros to the right of xyz is k-m, for k>m. e.g, 2300 is HAPPY (by proposition 1), so 0230=230 is also HAPPY.
case 2: if xyz is HAPPY, then by proposition 1, xyz*10^k =xyz000...0 is HAPPY, so permutin we have, x000...0y00..0z000...0

PROPOSITION 3:

proof: obvious

PROPOSITION 4:

proof; the number 111...000, with m ones and n zeros follows the undelisted sequenc;
1²+1²+1²+...+0²+0²+0²=1+1+1+...+0+0+0=m, hence if m is HAPPY, the number is also happy

COROLLARY 3: there are infinitely many binary numbers

proof: from above, for any HAPPY number m, there exist a binary number containin m ones, or equivalently, there exist a one to one correspondenc between HAPPY numbers and binary numbers, but from COROLLARY 1 above, there are infinitely many HAPPY numbers, and the result follows.

PROPOSITION 5:

proof: obviously follows from 3.)

PROPOSITION 6:

proof: let xy and ab be two numbers such that x+a<=9, y+b<=9 , suppose both numbers are HAPPY, then x²+y²=pqrs, is HAPPY and a²+b²=uvwz is also HAPPY.
Now xy+ab=(x+a)(y+b) so the sequenc of HAPPINESS for the sum is; (x+a)²+(y+b)²=x²+a²+2ax+y²+b²+2by=x²+y²+a²+b²+2(ax+by), now the number 2(ax+by) is not alwaz HAPPY (why?, because the product of HAPPY or SAD numbers is not alwaz happy, by proposition 3 above), and the result follows. In particular, if ab and xy are HAPPY then ab+xy is HAPPY iff 2(ax+by) is HAPPY. The same reasonin goes for a subtraction.
PROPOSITIO 7: Loading...

Anyone who proves the last gets a reward from me...promise...
Hey!, my proofs above are open 2 criticism

hmm, greetings to you all

Laplacian:
proofs:

PROPOSITION 1:

proof: let xy be a two digit HAPPY number, then the followin sequenc holds;
x²+y²=abcd

a²+b²+c²+d²=mnpqsr
.....
.....
..... =10000....0

Suppose we have xy*10^k for some integer k, then
xy*10^k=xy00000...0 with the zeros in k places, hence
x²+y²+0²+0²+...+0²=abcd, and the above sequenc is repeatd and the result follows.

COROLLARY 1: there are infinitely many HAPPY numbers.

Proof; because k is a natural number and there are infinitely many natural numbers, the result follows

PROPOSITION 2:

proof: let xyz be HAPPY, then
x²+y²+z²=abcd, where abcd is HAPPY, now the permutation yxz is shows that y²+x²+z²=x²+y²+z²=abcd, {because addition is cummutative}, and the result follows..

COROLLARY 2: if any number of zeros is either struck out or fixed in between the digits of a HAPPY number, the new number formed is also HAPPY.

Proof:
case 1: let the number be
xyz0000...0 havin k zeros, then by the above proposition, the number can be permuted to take m zeros to the left of the natural numbers, i.e
000...0xyz000..0=xyz000...0, where the zeros to the right of xyz is k-m, for k>m. e.g, 2300 is HAPPY (by proposition 1), so 0230=230 is also HAPPY.
case 2: if xyz is HAPPY, then by proposition 1, xyz*10^k =xyz000...0 is HAPPY, so permutin we have, x000...0y00..0z000...0

PROPOSITION 3:

proof: obvious

PROPOSITION 4:

proof; the number 111...000, with m ones and n zeros follows the undelisted sequenc;
1²+1²+1²+...+0²+0²+0²=1+1+1+...+0+0+0=m, hence if m is HAPPY, the number is also happy

COROLLARY 3: there are infinitely many binary numbers

proof: from above, for any HAPPY number m, there exist a binary number containin m ones, or equivalently, there exist a one to one correspondenc between HAPPY numbers and binary numbers, but from COROLLARY 1 above, there are infinitely many HAPPY numbers, and the result follows.

PROPOSITION 5:

proof: obviously follows from 3.)

PROPOSITION 6:

proof: let xy and ab be two numbers such that x+a<=9, y+b<=9 , suppose both numbers are HAPPY, then x²+y²=pqrs, is HAPPY and a²+b²=uvwz is also HAPPY.
Now xy+ab=(x+a)(y+b) so the sequenc of HAPPINESS for the sum is; (x+a)²+(y+b)²=x²+a²+2ax+y²+b²+2by=x²+y²+a²+b²+2(ax+by), now the number 2(ax+by) is not alwaz HAPPY (why?, because the product of HAPPY or SAD numbers is not alwaz happy, by proposition 3 above), and the result follows. In particular, if ab and xy are HAPPY then ab+xy is HAPPY iff 2(ax+by) is HAPPY. The same reasonin goes for a subtraction.
PROPOSITIO 7: Loading...

Anyone who proves the last gets a reward from me...promise...
Hey!, my proofs above are open 2 criticism

GOOD PROOF

Laplacian:
proofs:

PROPOSITION 1:

proof: let xy be a two digit HAPPY number, then the followin sequenc holds;
x²+y²=abcd

a²+b²+c²+d²=mnpqsr
.....
.....
..... =10000....0

Suppose we have xy*10^k for some integer k, then
xy*10^k=xy00000...0 with the zeros in k places, hence
x²+y²+0²+0²+...+0²=abcd, and the above sequenc is repeatd and the result follows.

COROLLARY 1: there are infinitely many HAPPY numbers.

Proof; because k is a natural number and there are infinitely many natural numbers, the result follows

PROPOSITION 2:

proof: let xyz be HAPPY, then
x²+y²+z²=abcd, where abcd is HAPPY, now the permutation yxz is shows that y²+x²+z²=x²+y²+z²=abcd, {because addition is cummutative}, and the result follows..

COROLLARY 2: if any number of zeros is either struck out or fixed in between the digits of a HAPPY number, the new number formed is also HAPPY.

Proof:
case 1: let the number be
xyz0000...0 havin k zeros, then by the above proposition, the number can be permuted to take m zeros to the left of the natural numbers, i.e
000...0xyz000..0=xyz000...0, where the zeros to the right of xyz is k-m, for k>m. e.g, 2300 is HAPPY (by proposition 1), so 0230=230 is also HAPPY.
case 2: if xyz is HAPPY, then by proposition 1, xyz*10^k =xyz000...0 is HAPPY, so permutin we have, x000...0y00..0z000...0

PROPOSITION 3:

proof: obvious

PROPOSITION 4:

proof; the number 111...000, with m ones and n zeros follows the undelisted sequenc;
1²+1²+1²+...+0²+0²+0²=1+1+1+...+0+0+0=m, hence if m is HAPPY, the number is also happy

COROLLARY 3: there are infinitely many binary numbers

proof: from above, for any HAPPY number m, there exist a binary number containin m ones, or equivalently, there exist a one to one correspondenc between HAPPY numbers and binary numbers, but from COROLLARY 1 above, there are infinitely many HAPPY numbers, and the result follows.

PROPOSITION 5:

proof: obviously follows from 3.)

PROPOSITION 6:

proof: let xy and ab be two numbers such that x+a<=9, y+b<=9 , suppose both numbers are HAPPY, then x²+y²=pqrs, is HAPPY and a²+b²=uvwz is also HAPPY.
Now xy+ab=(x+a)(y+b) so the sequenc of HAPPINESS for the sum is; (x+a)²+(y+b)²=x²+a²+2ax+y²+b²+2by=x²+y²+a²+b²+2(ax+by), now the number 2(ax+by) is not alwaz HAPPY (why?, because the product of HAPPY or SAD numbers is not alwaz happy, by proposition 3 above), and the result follows. In particular, if ab and xy are HAPPY then ab+xy is HAPPY iff 2(ax+by) is HAPPY. The same reasonin goes for a subtraction.
PROPOSITIO 7: Loading...

Anyone who proves the last gets a reward from me...promise...
Hey!, my proofs above are open 2 criticism

Nice one!

PROVE THE FIRST , SECOND,AND THIRD ISOMORPHISM THOREM ( ABSTRACT ALGEBRA)
HINT: SHOW THAT A FUNCTION IS ONE-ONE AND ALSO SHOW THAT IS A UNTO .MAPPING.
HAPPY DAY TO ALL THE MATHZ GURU.!

Chai! I don't know where to start or continue from. I was banned for over 3 days till now. He de pain me well well for some reply whe I be want give but no fit give right away. He de paine me well wellu o ga ga.
I'm happy to be back.

Humphrey77: -if x is equalto (9-0)(9-1)(9-2)(9-3)...(9-n) for all n is an integer x is ? @ jackpot please kindly give me the solution to the above question with in 10 min

Humphrey77: HERE IT
SOLUTION
(9-0)(9-1)(9-2)(9-3)(9-4)(9-5)(9-6)(9-7)(9-(9-9)(9-10)... AT THE POINT (9-9) WE HAVE ZERO . CLEARLY , (: THE PRODUCT EXPANSION AS IT VAULE EQUAL TO ZERO WITH WHICHTHE BEAKING POINTEXIST AT (9-9) )

Humphrey77: THE EXPRESSION (9-0)(9-1)(9-2)(9-3)(9-4)... IS EQAUL TO ZERO

HAPPY

Yes you are right but only for when n is greater than or equal to 9
What about when n < 9
For n = 0, X = 9 = 9 * 8! / 8!
n = 1, X = 9 * 8 = 9 * 8 * 7! / 7!
n = 2, X = 9 * 8 * 7 = 9 * 8 * 7 * 6! / 6!
This is the same as X = 9! / (8 - n)!

Hence X = (9-0)(9-1)(9-2)(9-3)...(9-n) can be written as
For 0 < n < 9, X = 9! / (8 - n)!
n greater than or equal to 9, X = 0
Well as for n < 0, that one pass me o.

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......

Laplacian:
...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...

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

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

I guess I know why @Laplacian said there are infinite number of triples that agree with this condition. From what I observed, I also agree with him.

1, 8, 15
4, 30, 56
15, 112, 127
56, 418, 780
2911, 21728, 40545

Take a look again, you'll notice that you can form another triple from the previous one because, half of the second term in the first triple is the first term in the next one while twice the 3rd term is the same as the second term in the next term. Since you already know the first and second term, you can get the third term. Following this trend, you can get an infinite number of triples.

This can be written as
X, 2Y, Z
Y, 2Z, 4Z - 2Y
Z, 8Z - 4Y, 15Z - 8Y
4Z - 2Y, 30Z - 16Y, 56Z - 30Y
15Z - 8Y, 112Z - 60Y, 209Z - 112Y
56Z - 30Y, 418Z -224Y,
209Z - 112Y,

^^i was equally unable to post 4d pass 72hrs ,,for unknown reasons.say x.
hm. oga richiez pls wot happend again?

benbuks: ^^i was equally unable to post 4d pass 72hrs ,,for unknown reasons.say x.
hm. oga richiez pls wot happend again?

Mine was because of what I posted. The message was repeated and one of the repeated was made all bold. I wanted to modify right away only to find out that I've been blocked from posting on the education section thanks to antispam bot. It was a lesson learnt the hard way.

akpos4uall:





Yes you are right but only for when n is greater than or equal to 9
What about when n < 9
For n = 0, X = 9 = 9 * 8! / 8!
n = 1, X = 9 * 8 = 9 * 8 * 7! / 7!
n = 2, X = 9 * 8 * 7 = 9 * 8 * 7 * 6! / 6!
This is the same as X = 9! / (8 - n)!

Hence X = (9-0)(9-1)(9-2)(9-3)...(9-n) can be written as
For 0 < n < 9, X = 9! / (8 - n)!
n greater than or equal to 9, X = 0
Well as for n < 0, that one pass me o.

why are you disturbing your self on (9-0)(9-1)(9-2)(9-3)(9-4)(9-5)(9-6)(9-7)(9-(9-9)(9-10)(9-11)... at the point (9-9) We have allthe product affected , which result to ZERO.
HAPPY

akpos4uall:








I guess I know why @Laplacian said there are infinite number of triples that agree with this condition. From what I observed, I also agree with him.

1, 8, 15
4, 30, 56
15, 112, 127
56, 418, 780
2911, 21728, 40545

Take a look again, you'll notice that you can form another triple from the previous one because, half of the second term in the first triple is the first term in the next one while twice the 3rd term is the same as the second term in the next term. Since you already know the first and second term, you can get the third term. Following this trend, you can get an infinite number of triples.

This can be written as
X, 2Y, Z
Y, 2Z, 4Z - 2Y
Z, 8Z - 4Y, 15Z - 8Y
4Z - 2Y, 30Z - 16Y, 56Z - 30Y
15Z - 8Y, 112Z - 60Y, 209Z - 112Y
56Z - 30Y, 418Z -224Y,
209Z - 112Y,

take a look at the third sequence it is not an AP OK.

akpos4uall:








I guess I know why @Laplacian said there are infinite number of triples that agree with this condition. From what I observed, I also agree with him.

1, 8, 15
4, 30, 56
15, 112, 127
56, 418, 780
2911, 21728, 40545

Take a look again, you'll notice that you can form another triple from the previous one because, half of the second term in the first triple is the first term in the next one while twice the 3rd term is the same as the second term in the next term. Since you already know the first and second term, you can get the third term. Following this trend, you can get an infinite number of triples.

This can be written as
X, 2Y, Z
Y, 2Z, 4Z - 2Y
Z, 8Z - 4Y, 15Z - 8Y
4Z - 2Y, 30Z - 16Y, 56Z - 30Y
15Z - 8Y, 112Z - 60Y, 209Z - 112Y
56Z - 30Y, 418Z -224Y,
209Z - 112Y,

take a look at the third sequence it is not an AP OK.
it ( 15,112,209) ,
:*@ HAPPY/

jackpot: that one na baby formula na. A "very obvious" one

Lemme share another Pythagorean formula which I have observed for you:

Proposition
Let k be any number at all. Then, the numbers:
k²-1, 2k and k²+1
are Pythagorean triples.

Proof
It follows from the application of the pythagorean theorem. QED.

Application
For k=2. . . . . . .3, 4, 5
k=3. . . . . . . 8, 6, 10
k=4. . . . . . . 15, 8, 17
k=5. . . . . . . 24, 10, 26
k=6. . . . . . . 35, 12, 37
k=7. . . . . . . 48, 14, 50.


Here is another one which I deduced from the above:

Corollary
Let k be an odd number. Then the numbers:
¹/₂(k²-1), k, ¹/₂(k²+1)
are Pythagorean triples.

Application
k=3. . . . . . . .4, 3, 5
k=5. . . . . . . .12, 5, 13
k=7. . . . . . . .24, 7, 25
k=9. . . . . . . .40, 9, 41

Propositions (jackpot 2013)
The following are pythagorean triples:
k^2-1, 2k, k^2+1
k^2-4, 4k, k^2+4
k^2-9, 6k, k^2+9
k^2-16, 8k, k^2+16

generally,
k^2-m^2, 2m, k^2+m^2

where 1<m<k and both k,m€ lN.


Other deductions are possible(think along the line the Corollary stated).


Gracias.

Nice and awesome. There is more to that. If you notice, with the given condition, the difference between 2 of the triples formed is usually 1. You can form much more as shown below;
From the theorem,
A² + B² = C²
A² = C² - B²
A² = (C + B)(C - B) ..... (1)

This can be interpreted as the product of the sum of two pythagorean triples and the difference between them is the same as the square of the third.

Let C + B = s ..... (2)
C - B = d ..... (3)
Substitute into equation (1) to get
A² = s*d ..... (4)
s = A² / d ...... (5)
Solving equations (2) & (3),
C = 0.5(s + d), B = 0.5(s - d)
substitute equation (5) into the above to get
C = 0.5 A² /d + d
C = (A² + d²) / 2d
similarly
B = (A² - d²) /2d
With this, we get
A, (A² - d²) /2d, (A² + d²) /2d

Using A = 12, several triples can be formed by changing the value of d
When d = 1, we'll get
12, (144 - 1) /2, (144 + 1) /2 = 12 , 71.5, 72.5
When d = 2, we'll get
12, (144 - 4) /4, (144 + 4) /4 = 12, 35, 37
When d = 3, we'll get
12, (144 - 9) /6, (144 + 9) /6 = 12, 22.5, 25.5
When d = 4, we'll get
12, (144 - 16)/8, (144 + 16)/8 = 12, 16, 20
When d = 6, we'll get
12, (144 - 36)/12, (144 + 36)/12 = 12, 9, 15
When d = 8, we'll get
12, (144 - 64)/ 16, (144 + 64)/16 = 12, 5, 13

[quote
author=akpos4uall]








I guess I know why @Laplacian said there are infinite number of triples
that agree with this condition. From what I observed, I also agree with
him.

1, 8, 15
4, 30, 56
15, 112, 127
56, 418, 780
2911, 21728, 40545

Take a look again, you'll notice that you can form another triple from
the previous one because, half of the second term in the first triple is
the first term in the next one while twice the 3rd term is the same as
the second term in the next term. Since you already know the first and
second term, you can get the third term. Following this trend, you can
get an infinite number of triples.

This can be written as
X, 2Y, Z
Y, 2Z, 4Z - 2Y
Z, 8Z - 4Y, 15Z - 8Y
4Z - 2Y, 30Z - 16Y, 56Z - 30Y
15Z - 8Y, 112Z - 60Y, 209Z - 112Y
56Z - 30Y, 418Z -224Y,
209Z - 112Y,
[/quote]
i luv ur observation..nice wrk!...if u get d proof, then u hav found another algorithm 4 Pell's eqn

Humphrey77: take a look at the third sequence it is not an AP OK.
it ( 15,112,209) ,
:*@ HAPPY/

Let me make it clearer;
From what I posted earlier,
X, 2Y, Z
Y, 2Z, 4Z - 2Y
Z, 8Z - 4Y, 15Z - 8Y
4Z - 2Y, 30Z - 16Y, 56Z - 30Y
15Z - 8Y, 112Z - 60Y, 209Z - 112Y
56Z - 30Y, 418Z -224Y,
209Z - 112Y,
* I just found out that I made a mistake*
That changes everything. Substituting
X = 1, Y = 4, Z = 15
should give us the right values if not for my mistake

Here are the right values
1, 8, 15
4, 30, 56
15, 112, 209
56, 418, 780
209, 1560, 2911
780, 5822, 10864
2911, 21728, 40545
10864, 81090, 151316
It goes on and on

Like I posted earlier, half the 2nd term will give you the 1st term on the next triple while twice the 3rd term will give you the 2nd term on the next triple. Since you've gotten the 1st & 2nd term for another triple, the third term of the new triple can be gotten by multiplying 2 by the 2nd term then subtract the 1st term from it.

Laplacian:
i luv ur observation even though it has flaws...nice wrk!
1.)first, ur second sequence should b; y, 2z, 4z-y....so all subsequent sequences are wrong.
2.) secondly, if u make d above corrections, and fit in real numbers, d first four sequences 'll b correct but all subsequent ones do not satisfy d condition of d question... Tanks

I noticed the mistake while I was fixing in the values.
As for the subsequent ones not satisfying the condition of the question, I guess you made a mistake along the line. With the help of a spread-sheet, I have the square roots of the perfect squares formed from the initial condition as shown below
3, 4, 11
11, 15, 41
41, 56, 153
153, 209, 571
571, 780, 2131
2131, 2911, 7953
7953, 10864, 29681
29681, 40545, 110771
There is still a trend here also.
The 3rd term is the same as the 1st term for the next triple, the 2nd term of this new triple is the same as the sum of the 2nd and 3rd term of the previous triple while the 3rd one is the same as multiplying the 2nd term by 2 then add the 1st term to it.
Chai! e no easy to use words explain wetin mathematical equation go describe for one line easily. I for just use variables make the picture de clearer.

akpos4uall:

I noticed the mistake while I was fixing in the values.
As for the subsequent ones not satisfying the condition of the question, I guess you made a mistake along the line. With the help of a spread-sheet, I have the square roots of the perfect squares formed from the initial condition as shown below
3, 4, 11
11, 15, 41
41, 56, 153
153, 209, 571
571, 780, 2131
2131, 2911, 7953
7953, 10864, 29681
29681, 40545, 110771
There is still a trend here also.
The 3rd term is the same as the 1st term for the next triple, the 2nd term of this new triple is the same as the sum of the 2nd and 3rd term of the previous triple while the 3rd one is the same as multiplying the 2nd term by 2 then add the 1st term to it.
Chai! e no easy to use words explain wetin mathematical equation go describe for one line easily. I for just use variables make the picture de clearer.

...yea, i 've corrected that, u posted ur comment just as i was about to modify it... Ur result is very impressive...pls i need complete details (very elaborate) on how u generated those programs on ur PC...are u on Facebook?...

Laplacian:
...yea, i 've corrected that, u posted ur comment just as i was about to modify it... Ur result is very impressive...pls i need complete details (very elaborate) on how u generated those programs on ur PC...are u on Facebook?...

@ akpos ; your sequence did not satisfy laplacian question! that is three non- zero integers are in AP, the product of any two when increase by one is a perfect square.
:*HAPPY

Consider the digit from 1to 9 , select any three digit form (1,2,3,4,5,6,7,8,9) with out repetition the sum of the digit's will give 15 in the column , diagonal and row
HAPPY!

Humphrey77: @ akpos ; your sequence did not satisfy laplacian question! that is three non- zero integers are in AP, the product of any two when increase by one is a perfect square.
:*HAPPY

Let me use equations to make it clearer to you. Hope you know that I have two sets? The first one agrees with the question while the second one is the square roots of the perfect squares from the question.

According to the question, if (a, b, c) agree with the condition,
ab + 1 = X²
ac + 1 = Y²
bc + 1 = Z²
(a, b, c) & (X, Y, Z) are the two sets that I posted. Let M_n = (a_n, b_n, c_n) & P_n = (X_n, Y_n, Z_n)

We were able to get M₁ = (1, 8, 15) but I noticed that the subsequent ones that were gotten, had a relationship with M₁ as follows
M₁ = (a₁, b₁, c₁)
M₂ = (a₂, b₂, c₂)
M₃ = (a₃, b₃, c₃
Where a₂ = 0.5b₁, b₂ = 2c₁, c₂ = 2b₂ - a₂
Likewise a₃ = 0.5b₂, b₃ = 2c₂, c₃ = 2b₃ - a₃
Generally
M₁ = (a₁, b₁, c₁) = (1, 8, 15)
M_(n+1) = (0.5b_n, 2c_n, 2b_(n+1) - a_(n+1))
As for P_n
P_(n+1) = (Z_n, Z_n + Y_n, 2Y_(n+1) + X_(n+1)
P₁ = (X₁, Y₁, Z₁) = (3, 4, 11)


M₁ = 1, 8, 15
M₂ = 4, 30, 56
M₃ = 15, 112, 209
M₄ = 56, 418, 780
M₅ = 209, 1560, 2911
M₆ = 780, 5822, 10864
M₇ = 2911, 21728, 40545
M₈ = 10864, 81090, 151316

P₁ = 3, 4, 11
P₂ = 11, 15, 41
P₃ = 41, 56, 153
P₄ = 153, 209, 571
P₅ = 571, 780, 2131
P₆ = 2131, 2911, 7953
P₇ = 7953, 10864, 29681
P₈ = 29681, 40545, 110771

akpos4uall:

Let me use equations to make it clearer to you. Hope you know that I have two sets? The first one agrees with the question while the second one is the square roots of the perfect squares from the question.

According to the question, if (a, b, c) agree with the condition,
ab + 1 = X²
ac + 1 = Y²
bc + 1 = Z²
(a, b, c) & (X, Y, Z) are the two sets that I posted. Let M_n = (a_n, b_n, c_n) & P_n = (X_n, Y_n, Z_n)

We were able to get M₁ = (1, 8, 15) but I noticed that the subsequent ones that were gotten, had a relationship with M₁ as follows
M₁ = (a₁, b₁, c₁)
M₂ = (a₂, b₂, c₂)
M₃ = (a₃, b₃, c₃
Where a₂ = 0.5b₁, b₂ = 2c₁, c₂ = 2b₂ - a₂
Likewise a₃ = 0.5b₂, b₃ = 2c₂, c₃ = 2b₃ - a₃
Generally
M₁ = (a₁, b₁, c₁) = (1, 8, 15)
M_(n+1) = (0.5b_n, 2c_n, 2b_(n+1) - a_(n+1))
As for P_n
P_(n+1) = (Z_n, Z_n + Y_n, 2Y_(n+1) + X_(n+1)
P₁ = (X₁, Y₁, Z₁) = (3, 4, 11)


M₁ = 1, 8, 15
M₂ = 4, 30, 56
M₃ = 15, 112, 209
M₄ = 56, 418, 780
M₅ = 209, 1560, 2911
M₆ = 780, 5822, 10864
M₇ = 2911, 21728, 40545
M₈ = 10864, 81090, 151316

P₁ = 3, 4, 11
P₂ = 11, 15, 41
P₃ = 41, 56, 153
P₄ = 153, 209, 571
P₅ = 571, 780, 2131
P₆ = 2131, 2911, 7953
P₇ = 7953, 10864, 29681
P₈ = 29681, 40545, 110771

9ICE-ONE

Humphrey77: Consider the digit from 1to 9 , select any three digit form (1,2,3,4,5,6,7,8,9) with out repetition the sum of the digit's will give 15 in the column , diagonal and row
HAPPY!

This is known as the magic square. I came across this in MATLAB. A special type of matrix known as magic square matrix http://en.m.wikipedia.org/wiki/Magic_square
I don't know how it is solved sha but this is the square for a 3 by 3
2 7 6
9 5 1
4 3 8

SOLVE FOR h in the equation : 6(raise to power h) -31(2)raise to power h+32 is equal to zero ( HINT : USE LINEAR APPROX)
:* HAPPY!

akpos4uall:

This is known as the magic square. I came across this in MATLAB. A special type of matrix known as magic square matrix http://en.m.wikipedia.org/wiki/Magic_square
I don't know how it is solved sha but this is the square for a 3 by 3
2 7 6
9 5 1
4 3 8

GOOD IT CAN SO BE ;
8 3 4
1 5 9
6 7 2


HAPPY

Have gat two pending questions on laplace transform. Any attempt to it will be so mush appreciated..lets warm up with this

1. Show that:
Sin(45-x)/cos2x=sec(45-x)/2

2. If dx=-_/(100-y²)dy/y
show that x=-_/(100-y²) + 10 ln[10+_/(100-y²)/y] +c

pls the solution to Q2 is very important and really needed...i will start giving my full contributions in no time.Any atempt to the above is welcomed...

This is my last post for this year. I wish you all a merry xmas and a prosperous new year.

Catch y'all next year