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rajeshmarndi
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All composite number comprised of atleast two factor e.g
57 = 3 * 19 and 1 * 57
14927= 11*1357 = 23*649 = 59*253 = 1*14927
We can get to know 57 is a composite number , as the factor 3*19 is achieved, as below, from 11^2. The first factor reduces and the 2nd factor increases, till first factor is 1.
11*11=121
9*13=117
7*15=105
5*17=85
3*19=57
1*21=21
Note all these composite no. that descend from 11^2 are square distance from 11^2 as below.
121-117=4= 2^2
121-105=16 = 4^2
121-85=36 = 6^2
121-57=64 = 8^2
121-21= 100 = 10^2
Note 1*21 factor has always a difference of 1, here, 11-10=1 (i.e of 11^2 & 10^2). So it can always be eliminated.This is true for all composite number. So, each Composite number has a specific square(higher than the test number N) from where it descend, like in the above example.So to find whether a number N is a composite or not. One only need to see if any of the higher squares are square distance to N.
Say a test number N=1656337 . Square next to it is , integer(√N)+ 1
which is,
= integer (√1656337) ) + 1 = integer ( 1286.987...) ) + 1 = 1286 + 1 = 1287.
So, square next to N is 1287^2.
We can formulate distances to next squares which will be quadratic in nature. For the above number N, it will be,
x^2 + 1287*2*x + (1287^2-N)
(this is the distances to all the higher square to N. Starting with x=0)
x^2 + 2574x + 32 = m^2 ?
So, if the above quadratic expression generates an perfect square (integer). Then N is a composite number. Remember 1*N also comes under a specific squares, so every such equation will definitely have one solution for 1*N factor.
So we need to see if the quadratic expression generates atleast two solution. If yes then definitely N is a composite number.
But I am not able to determine, how to say such a quadratic will generate perfect squares or not.For the above test number N. Since I know 1656337 is a composite number and 1217 and 1361 are its factor.
In that case, we can find the squares where its distance comes to be a square.
(1217 + 1361) /2 = 1289
So we get from 1289^2, 1217*1361 has descended.
so, 1289^2 - 1656337 = 72^2
i.e
x^2 + 2574x + 32 = 72^2
so we definitely get at x=2 the quadratic is a perfect square
Similarly for factor 1*1656337, it always descend from
integer(1656337/2) + 1 = integer(828168.5)+1 = 828168+1 = 828169 square
i.e
x^2 + 2574x + 32 = 828169^2
and x for this, will always be
x= [(1287-1)^2 - 32 ] / 2 = 826882
So,
x^2 + 2574x + 32 = m^2
has two solution.
i.e N(1656337) has two square at x= 2 & x = 826882 and therefore is a composite number.
Please comment.
Thank you.
57 = 3 * 19 and 1 * 57
14927= 11*1357 = 23*649 = 59*253 = 1*14927
We can get to know 57 is a composite number , as the factor 3*19 is achieved, as below, from 11^2. The first factor reduces and the 2nd factor increases, till first factor is 1.
11*11=121
9*13=117
7*15=105
5*17=85
3*19=57
1*21=21
Note all these composite no. that descend from 11^2 are square distance from 11^2 as below.
121-117=4= 2^2
121-105=16 = 4^2
121-85=36 = 6^2
121-57=64 = 8^2
121-21= 100 = 10^2
Note 1*21 factor has always a difference of 1, here, 11-10=1 (i.e of 11^2 & 10^2). So it can always be eliminated.This is true for all composite number. So, each Composite number has a specific square(higher than the test number N) from where it descend, like in the above example.So to find whether a number N is a composite or not. One only need to see if any of the higher squares are square distance to N.
Say a test number N=1656337 . Square next to it is , integer(√N)+ 1
which is,
= integer (√1656337) ) + 1 = integer ( 1286.987...) ) + 1 = 1286 + 1 = 1287.
So, square next to N is 1287^2.
We can formulate distances to next squares which will be quadratic in nature. For the above number N, it will be,
x^2 + 1287*2*x + (1287^2-N)
(this is the distances to all the higher square to N. Starting with x=0)
x^2 + 2574x + 32 = m^2 ?
So, if the above quadratic expression generates an perfect square (integer). Then N is a composite number. Remember 1*N also comes under a specific squares, so every such equation will definitely have one solution for 1*N factor.
So we need to see if the quadratic expression generates atleast two solution. If yes then definitely N is a composite number.
But I am not able to determine, how to say such a quadratic will generate perfect squares or not.For the above test number N. Since I know 1656337 is a composite number and 1217 and 1361 are its factor.
In that case, we can find the squares where its distance comes to be a square.
(1217 + 1361) /2 = 1289
So we get from 1289^2, 1217*1361 has descended.
so, 1289^2 - 1656337 = 72^2
i.e
x^2 + 2574x + 32 = 72^2
so we definitely get at x=2 the quadratic is a perfect square
Similarly for factor 1*1656337, it always descend from
integer(1656337/2) + 1 = integer(828168.5)+1 = 828168+1 = 828169 square
i.e
x^2 + 2574x + 32 = 828169^2
and x for this, will always be
x= [(1287-1)^2 - 32 ] / 2 = 826882
So,
x^2 + 2574x + 32 = m^2
has two solution.
i.e N(1656337) has two square at x= 2 & x = 826882 and therefore is a composite number.
Please comment.
Thank you.