Perfect square quadratic

Number Theory Level 4

Let n 2 − 3 n − 126 n^2-3n-126 be a perfect square for certain { n ∣ n ∈ Z } \{n\text{ }|\text{ } n\in\mathbb{Z}\} . What is the sum of all distinct, possible values of n n ?


The answer is 12.

Aaron Tsai by Aaron Tsai , 94, USA

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1 solution

Aaron Tsai
Jun 5, 2016

Let n 2 − 3 n − 126 = k 2 n^2-3n-126=k^2 for some integer k k . Multiplying both sides by 4 4 , we have

4 n 2 − 12 n − 504 = 4 k 2 4n^2-12n-504=4k^2

Completing the square, we have

4 n 2 − 12 n + 9 − 513 = 4 k 2 4n^2-12n+9-513=4k^2

( 2 n ) 2 − 2 ( 2 n ) ( 3 ) + 3 2 − 513 = 4 k 2 (2n)^2-2(2n)(3)+3^2-513=4k^2

( 2 n − 3 ) 2 − 513 = 4 k 2 (2n-3)^2-513=4k^2

( 2 n − 3 ) 2 − ( 2 k ) 2 = 512 (2n-3)^2-(2k)^2=512

Using difference of squares,

( 2 n − 3 + 2 k ) ( 2 n − 3 − 2 k ) = 513 (2n-3+2k)(2n-3-2k)=513

Let a = 2 n − 3 + 2 k a=2n-3+2k and b = 2 n − 3 − 2 k b=2n-3-2k . Then a b = 513 ab=513 . Note that since n n and k k are integers, a a and b b must be integers as well. (Unordered) Pairs of integers that multiply to 513 513 are { 1 , 513 } , { 3 , 171 } , { 9 , 57 } , { 19 , 27 } \{1,513\}, \{3, 171\}, \{9,57\}, \{19,27\} and their opposites.

Adding a a and b b together, we have

4 n − 6 = a + b 4n-6=a+b

n = a + b + 6 4 n=\dfrac{a+b+6}{4}

Now we just have to evaluate n n for each pair of factors (and their opposites).

For { 1 , 513 } \{1,513\} , n = 1 + 513 + 6 4 = 130 n = \dfrac{1 + 513 + 6}{4} = 130

For { 3 , 171 } \{3,171\} , n = 3 + 171 + 6 4 = 45 n = \dfrac{3 + 171 + 6}{4} = 45

For { 9 , 57 } \{9,57\} , n = 9 + 57 + 6 4 = 18 n = \dfrac{9 + 57 + 6}{4} = 18

For { 19 , 27 } \{19,27\} , n = 19 + 27 + 6 4 = 13 n = \dfrac{19 + 27 + 6}{4} = 13

For { − 1 , − 513 } \{-1,-513\} , n = − 1 − 513 + 6 4 = − 127 n = \dfrac{-1 - 513 + 6}{4} = -127

For { − 3 , − 171 } \{-3, -171\} , n = − 3 − 171 + 6 4 = − 42 n = \dfrac{-3 - 171 + 6}{4} = -42

For { − 9 , − 57 } \{-9, -57\} , n = − 9 − 57 + 6 4 = − 15 n = \dfrac{-9 - 57 + 6}{4} = -15

For { − 19 , − 27 } \{-19, -27\} , n = − 19 − 27 + 6 4 = − 10 n = \dfrac{-19 - 27 + 6}{4} = -10

Thus, our answer is 130 + 45 + 18 + 13 − 127 − 42 − 15 − 10 = 12 130 + 45 + 18 + 13 - 127 - 42 - 15 - 10 = \boxed{12}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Your method is beautiful. You have used a nice method to avoid fractions. I give below your solution with some modifications, retaining all your variables and explanations.

T h e e q u a t i o n i s ( n − 3 2 ) 2 − 513 4 = k 2 , { k ∣ k ∈ Z } . T o a v o i d f r a c t i o n s m u l t i p l y b y 4. ∴ ( 2 n − 3 ) 2 − 4 k 2 = 513 L e t a = 2 n − 3 + 2 k a n d b = 2 n − 3 − 2 k . T h e n a b = 513. N o t e t h a t s i n c e n a n d k a r e i n t e g e r s , a a n d b m u s t b e i n t e g e r s a s w e l l . 513 = 3 3 ∗ 19. ⟹ ( 3 + 1 ) ( 1 + 1 ) = 8 f a c t o r s . I n p a i r s 4 p a i r s o f + t i v e a n d − t i v e n . A d d i n g a a n d b t o g e t h e r , f o r + t i v e , a n d − t i v e f a c t o r s w e h a v e , 4 n + − 6 = a + b g i v e s n + = a + b + 6 4 . 4 n − − 6 = − a − b g i v e s n − = a + b − 6 4 = − n + + 3. S o t h e p a i r , n + + n − = 3. S i n c e t h e r e a r e f o u r p a i r s , s u m o f a l l n = 4 ∗ ( + 3 ) = 12. The~ equation~ is ~\Big(n-\dfrac 3 2\Big)^2-\dfrac{513} 4=k^2,~\{k\text{ }|\text{ } k\in\mathbb{Z}\}.\\ ~~~\\ To ~avoid~fractions~multiply~by~ 4.~\therefore~(2n-3)^2-4k^2=513\\ ~~~\\ Let~a=2n-3+2k~ and~ b=2n-3-2k.~ Then~ab=513.~\\ ~~~\\ \color{#3D99F6}{ Note~ that~ since~ n ~and~k~ are~ integers,~a~ and~b~ must~ be~ integers~ as~ well.}\\ ~~~\\ 513=3^3*19. ~\implies~(3+1)(1+1)=8~factors.~~In~ pairs ~4~pairs~of~+tive~and~-tive~n.\\ ~~~\\ Adding~ a~ and~b~ together,~for~+tive,~and~-tive~factors~ we~ have,\\ ~~~\\ 4n_{+}-6=a+b~gives~n_{+}=\dfrac{a+b+6}4.\\ ~~~\\ 4n_{-}-6=-a-b~gives~n_{-}=\dfrac{a+b-6}4=-n_{+}+3.\\ ~~~\\ So~~the~pair, ~n_{+}~+~n_{-}=\color{#3D99F6}{3}.\\ ~~~\\ Since~there~are~four~pairs,~sum~of~ all~ n= 4*(+3)=\Large~~\color{#D61F06}{12}.\\

N o t e : − T h e ( U n o r d e r e d ) P a i r s f a c t o r s { X , Y } a r e : − A = { 1 , 513 } , B = { 3 , 171 } , C = { 9 , 57 } , D = { 19 , 27 } , − A , − B , − C , − D . ⟹ A + ( − A ) + B + ( − B ) + C + ( − C ) + D + ( − D ) . F o r ± A n + = 1 + 513 + 6 4 = 130 , n − = − 127. F o r ± B n + = 3 + 171 + 6 4 = 45 , n − = − 47. F o r ± C n + = 9 + 57 + 6 4 = 18 , n − = − 15. F o r ± D n + = 19 + 27 + 6 4 = 13 , n − = − 10. S u m a l l n = 130 − 127 + 45 − 42 + 18 − 15 + 13 − 10 = 12. Note:-\\ The~(Unordered)~ Pairs~ factors~\{X,~Y\}~~are:-~~A=\{1,~513\}~,~B=\{3,~171\}~,~C=\{9,~57\}~,~D=\{19,~27\},~-A,~-B,~-C,~-D.\\ \implies~~A+(-A)~~~+~~~B+(-B)~~~+~~~ C+(-C)~~~+~~~D+(-D).\\ For~\pm A~~n_{+}=\dfrac{1+513+6} 4=130,~~~ n_{-}=-127.\\ For~\pm B~~n_{+}=\dfrac{3+171+6} 4=~45,~~~ n_{-}=-~47.\\ For~\pm C~~n_{+}=\dfrac{9+57+6} 4=~18,~~~ n_{-}=-~15.\\ For~\pm D~~n_{+}=\dfrac{19+27+6} 4=13,~~~ n_{-}=-~10.\\ Sum~all~n~=130-127~~~+~~~45-42~~~+~~~18-15~~~+~~~13-10=12.

Niranjan Khanderia - 3 years, 2 months ago

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