https://doi.org/10.22214/ijraset.2023.49490
11 III
2023
March
ISSN: 2321-9653; IC Value: 45.98; SJ Impact Factor: 7.538
Volume 11 Issue III Mar 2023- Available at www.ijraset.com
Integral Solutions of the Ternion Quadratic Equation
Gowri Shankari A1 , Sangavi S2
1Assistant Professor, PG and Research Department of Mathematics, 2PG Student, Cauvery College for Women (Autonomous), Trichy-18, (TamilNadu) India
Abstract: In order to find its non zero unique integral solutions for the quadratic diophantine equation with three unknowns given by is analysed. The equation under consideration exhibits multiple patterns of solutions. The solutions are presented with a few fascinating aspects.
Keywords: Quadratic equation with three unknowns, integral solutions, polygonal numbers.
I. INTRODUCTION
The quadratic diophantine equation with three unknowns offers a numerous researching opportunities because of their range[1-3]. For quadratic equations containing three unknowns, one should specially refer [4-19]. This communication deals with yet another fascinating ternary quadratic equation 3 2 2 401s g a with three unknown factors that can be used to determine any one of an infinite numbers of non-zero integral solutions.
II. METHOD OF ANALYSIS
The ternion quadratic equation to be solved for its non-zero integral solution is 2 2 2 401s g a (1) various patterns of solutions of (1) are listed below
A. Pattern 1: Assume,
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2 2 2
401s g a
4,
A. Notations 2
n T n (Tetragonal number) 3) (4 10, n n T n (Decagonal number) 4) (5 12, n n T n (Dodecagonal number) 6) (7 16, n n T n ( Hexadecagonal number) 8) (9 20, n n T n (Icosagonal number)
2 2
1
9) (10 22, n n T n (Icosidigonal number) 10) (11 24, n n T n (Icositetragonal number) 12) (13 28, n n T n (Icosioctagonal number)
1) ( n n CS n (Centered square number)
2 n Gnon (Gnomonic number) 1 3 3 2 n n CH n (Centered hexagonal number)
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Where l and m are non-zero integers and write
Using (2) and (3) in (1), we get
Using factorization method, we have
If we analyse the positive and negative aspects, we
Equating the actual and fictitious elements either
PROPERTIES
B. Pattern 2
We write 401 in the form as
731 ©IJRASET: All Rights are Reserved | SJ Impact Factor 7.538 | ISRA Journal Impact Factor 7.894 | 2 2 ) , ( m l m l s s (2)
) )(20 (20 401 i i (3)
2 2 2 2 2 ) )( )(20 (20 m l i i g a
2 2 ) ( ) )( )(20 (20 ) )( ( im l im l i i ig a ig a
get 2) )( (20 im l i ig a (4) 2) )( (20 im l i ig a (5)
in (4)
we get 2 2 20 2 20 ) , ( m lm l m l a 2 2 40 ) , ( m lm l m l g 2 2 ) , ( m l m l s
or (5),
1) square perfect (1,1)isa 2 (1,1) a g 2) nastynumber (1,1)isa 3 (1,1) 2 (1,1) s a g 3) 0(mod27) 7 4 ,1) ( 40 ,1) ( 60 12, l l Gno T l a l s 4) 0(mod2) 25 2 ) (1, 31 ) (1, ) (1, 12, l l Gno T l s l g l a 5) 42) 0(mod 21 ,1) ( 19 ,1) ( ,1) ( lGno l s l a l g 6) 0 20 ) , ( 30 ) , ( 4, l T l l s l l a 7) 0 2 ) , ( 4, m T m m s 8) number perfect (1,1)isa
(1,1) (1,1) 20
a 9) Disariumnumber (1,1)isa 60 (1,1) 10 (1,1) s g a 10) number palindrome a (1,1)is 3 (1,1) (1,1) s g a
7
s g
) 20 )(1 20 (1 401 i i (6)
(6) and (2) in (1), we get 2 2 2 2 2 ) )( 20 )(1 20 (1 m l i i g a Using factorization method, we have 2 2 ) ( ) )( 20 )(1 20 (1 ) )( ( im l im l i i ig a ig a
aspects, we get 2) )( 20 (1 im l i ig a (7) 2) )( 20 (1 im l i ig a (8)
in (7) or (8), we get 2 2 40 ) , ( m lm l m l a
Where l and m are non-zero integers and write Using
If we analyse the positive and negative
Equating the actual and fictitious elements either
PROPERTIES
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C. Pattern 3
Observe that (1) is written as
which is equivalent to the system of double equation
By applying the cross multiplication method to solve (10), the relevant non-zero unique integral solution to (1) are obtained as
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1) square perfect (1,1)isa 38 (1,1) s a 2) number perrin a (1,1)is 20 (1,1) (1,1) s g a 3) 37) 0(mod 11 10 ,1) ( 39 ,1) ( ,1) ( l l Gno CS l s l g l a 4) 0(mod141) 23 ) (1, 59 ) (1, ) (1, 3 lGno l s l a l g 5) 22) 0(mod 20 ,1) ( ,1) ( mGno m a m s 6) 0 ) , ( 10 , 10 ) , ( l l s l l g l l a 7) 0 20 ) , ( 30 ) , ( 4, l T l l s l l a
number
s g
number
s a
g
8)
deficient (1,1)isa (1,1)
9)
palindrome (1,1)isa 31 (1,1)
10) nastynumber (1,1)isa 2 (1,1) s
2 2 2 2 400 s s g a (9) g s s a 20 s a g s 20 , 0
0 ) (20 s g a 0 ) 20 ( s g a (10)
2 2 20 2 20 ) , ( a 2 2 40 ) , ( g 2 2 ) , ( s PROPERTIES 1) happynumber (1,1)isa 2 (1,1) (1,1) 2 s g a 2) number silverback a (1,1)is 7 (1,1) s g 3) 511) 0(mod 501 ) (1, 5 ) (1, ) (1, 25 Gno s a g 4) 0(mod140) 47 2 ,1) ( 30 ,1) ( ,1) ( 2 Gno Star s x g 5) 0(mod146) 71 20 ) (1, 42 ) (1, ) (1, 2 Gno CH s a g 6) 0 36 ) , ( 2 ) , ( 4, T s g 7) 0 6 ) , ( 4 ) , ( 2 4, T s a 8) number perrin (1,1)isa 31 (1,1) 20 s a 9) number palindrome (1,1)isa (1,1) (1,1) s a g 10) digital perfect (1,1)isa 25 (1,1) 50 (1,1) 10 s a g number invariant
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Pattern 4 Equation (1) is written as 1 400 2 2 2 s g a (11) Write 1 as 229 ) 21 )(20 21 (20 1 i i (12) and write 400 as ) )(20 (20 400 i i (13) Using (2), (12) and (13) in (11), we get 2 2 2 2 2 29 ) 21 )(20 21 (20 ) )( )(20 (20 i i m l i i g a Employing the method of factorization the above equation is written as 2 2 2 29 ) 21 )(20 21 (20 ) ( ) )( )(20 (20 ) )( ( i i im l im l i i ig a ig a If we analyse the positive and negative aspects, we get 2) )( 21 )(20 (20 29 1 im l i i ig a (14) 2) )( 21 )(20 29(20 1 im l i i ig a (15) Equating the actual and fictitious elements either in (14) or (15), we get ) 379 880 (379 29 1 ) , ( 2 2 m lm l m l a ) 440 758 29(440 1 ) , ( 2 2 m lm l m l g Since our interest is on finding integer solution Let us take L l 29 and M m 29 2 2 379 880 379 ) , ( M LM L M L a 2 2 440 758 440 ) , ( M LM L M L g 2 2 29 29 ) , ( M L M L s PROPERTIES 1) number abudant (1,1)isa 16 (1,1) s a 2) number perrin (1,1)isa (1,1) 14 a s 3) L L Gno CS L s L a L g 1724 526 ,1) ( 19 ,1) ( ,1) ( 2 0(mod1248) 4) 0(mod955) 831 2 ) (1, 3 ) (1, ) (1, 28, L L Gno T L s L a L g 5) 0(mod2129) 706 133 ,1) ( ,1) ( 36 12, L L Gno T L a L s 6) 0(mod827) 859 10 ,1) ( ,1) ( ,1) ( 20, M M Gno T M s M a M g 7) 0(mod967) 832 ,1) ( 2 ) (1, ) (1, MGno M s M a M g 8) 0 4 ) , ( 13 ) , ( 4, L T L L s L L g 9) 0 52 ) , ( 3 ) , ( ) , ( 4, M T M M s M M g M M a 10) square perfect (1,1)isa 13 (1,1) s g
D.
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E. Pattern 5
Equation (1) is written as
Using (17) and (18) in (16) and applying the method
III. CONCLUSION
we have given numerous non-zero unique integral solutions patterns. To conclude, one can look for further options for solutions and their respective attributes among the various choices.
For the ternion quadratic equation
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1 401 2 2 2 a g s (16) Assume, 2 2 401 ) , ( m l m l a (17) Write 1 as 20) 401 20)( 401 ( 1 (18)
of
we get 20) 401 ( ) 401 ( ) 401 ( 2 m l g s Equating then rational and irrational factors, we get 2 2 401 ) , ( m l m l a 2 2 20 802 8020 ) , ( m lm l m l g 2 2 40 401 ) , ( m lm l m l s PROPERTIES 1) 0(mod8523) 299 2 ) (1, 25 ) (1, 7 ) (1, 2 10, l l Gno T l s l a l g 2) 0(mod6827) 391 2 ) (1, 2 ) (1, 24, Gnol T l a l g l 3) 0(mod380) 401 ,1) ( ,1) ( 20 mGno m g m a 4) 0(mod8410) 412 2 ) (1, ) (1, ) (1, 22, m m Gno T m s m g m a 5) 0(mod8009) 412 3 ) (1, ) (1, 16, m m Gno T m s m g 6) 0(mod180) 200 ,1) ( 10 ,1) ( 10 lGno l s l a 7) 0 38 ) , ( ) , ( 4, l T l l s l l a 8) 0 382 ) , ( 10 ) , ( ) , ( 4, m T m m s m m g m m a 9) 0 21 ,1) ( 19 ,1) ( ,1) ( mGno m s m a m g 10) square perfect a (1,1)representsa
factorization,
2 2 2 401s g a
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