You Could Count Them All.

Number Theory Level 5

How many pairs of integers ( a , b ) (a,b) , where 0 ≤ a < b ≤ 100 0\le a<b\le 100 , satisfy ( b 2 − a 2 ) ≡ 2 ( m o d 5 ) , \left( b^{2}-a^{2}\right) \equiv 2\pmod{5}, ( b 2 − a 2 ) ≡ 5 ( m o d 8 ) ? \left( b^{2}-a^{2}\right) \equiv 5\pmod{8}?


The answer is 100.

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Walker Anderson by Walker Anderson , 20, USA

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1 solution

L e t E x p = ( b 2 − a 2 ) T o f i n d E x p ≡ 2 ( m o d 5 ) ≡ 5 ( m o d 8 ) . Let~ Exp~ =~ (~b^{2} - a^{2}~) ~~~~~~~To~ find~~Exp~~≡~2~(mod~5) ≡~5~(mod~8). ~~~ F i r s t i s t r u e i f u n i t v a l u e o f E x p i s 2 o r 7. O N L Y 7 c a n s a t i s f y t h e s e c o n d . First~is~true~if~ unit~value~of~Exp~is~2~~or~~7.~~ONLY~7 ~can~satisfy ~the~second.~ \color\green{So~unit ~value~of~~(~b^{2} - a^{2}~)~ ~is~~7.}

~A~ square~of~an~ integer~~can~have~only ~\color\red{0,~1,~4,~5,~6~and~9~}~ ~~ at~its~unit~value.

7 c a n b e o b t a i n e d f r o m ( b 2 , a 2 ) b y ( 11 , 4 ) . ~~~~~~~~~~~~~~7~can~ be~obtained~from~~(b^{2}, a^{2}) ~~by~ (11, 4).

T h a t i s ( b , a ) = ( 1 o r 9 , 2 o r 8 ) = ( 1 , 2 ) , ( 1 , 8 ) a n d ( 9 , 2 ) , ( 9 , 8 ) . That~is~~(b,a)=~(1~or~9, 2~o~r8)=(1,2),(1,8)~~and~~(9,2),(9,8).~

L e t u n i t v a l u e o f b = p , u n i t v a l u e o f a = q . X , Y = 0 , 1 , . . . 9. Let~unit ~value~of~ b=p,~~~~~~~unit~value~of~a=q.~~~~~~~X,~Y=0,~1,~.~.~.~9. S o w e h a v e b = 10 X + p a n d a = 10 Y + q . ~~~~~~~~~~~~~~~~~~~~~~~So~we~have~~~~~b=10X+p~~~~~and~~~~~a=10Y +q. \color\green {\therefore~~ (~b^{2} - a^{2}~)~ =~ (~10X + p~)^{2} - (10Y + q)^{2} }

∴ 0 ( m o d 8 ) ≡ { ( 10 X + p ) 2 − ( 10 Y + q ) 2 − 5 } ( m o d 8 ) ~~~~~~~\therefore~~0~(mod~8)~≡\{ (~10X + p~)^{2} - (10Y + q)^{2}-5\}~~ (mod~8) ≡ { 100 X 2 + 20 X p − 100 Y 2 − 20 Y q + p 2 − q 2 − 5 } ( m o d 8 ) ~~~~~~~~~~~~≡\{ 100X^{2}+20Xp-100Y^{2}-20Yq+p^{2}-q^{2}-5\}~~ (mod~8) \color\green{ ≡\{4X^{2}+4Xp-4Y^{2}-4Yq+p^{2}-q^{2}-5\}~~(mod~8)..........(1)}

For(b,a)=\color\red{(1,2)}:~~~~~~p=1,~q=2,~~~~~~~~~(1)~~~~~ becomes~~~~~ ≡ { 4 X 2 + 4 X − 4 Y 2 − 8 Y q + 1 − 4 − 5 } ( m o d 8 ) ≡\{4X^{2}+4X-4Y^{2}-8Yq+1-4-5\}~(mod~8)~~ ≡ { ( 4 X 2 + 4 X − 4 Y 2 } ( m o d 8 ) ( 1 ) t r u e o n l y f o r e v e n Y ≡\{(4X^{2}+4X-4Y^{2}\}~(mod~8)~~(1)~true~only~for~even~Y

For(b,a)=\color\red{(1,8)}:~~~~~~p=1,~q=8,~~~~~~~~~(1) ~~~~~becomes~~~~ ≡ { 4 X 2 + 4 X − 4 Y 2 − 8 Y q + 1 − 64 − 5 } ( m o d 8 ) ≡\{4X^{2}+4X-4Y^{2}-8Yq+1-64-5\}~(mod~8)~~ ≡ ( 4 X 2 + 4 X − 4 Y 2 − 4 ) ( m o d 8 ) ( 1 ) t r u e o n l y f o r o d d Y ≡(4X^{2}+4X-4Y^{2}-4)~(mod~8)~~~(1)~true~only~for~odd~Y

S i m i l a r l y f o r b o t h ( b , a ) = ( 9 , 2 ) , ( 9 , 8 ) ( 1 ) b e c o m e s Similarly~ for~~ both~~(b,a)= (9,2),(9,8)~~~~~~~~~~~(1)~~~~~ becomes
≡ ( 4 X 2 + 4 X − 4 ) ( m o d 8 ) w h i c h i s N E V E R ≡ 0 ( m o d 8 ) ≡(4X^{2}+4X-4)~(mod~8)~~which~ is~ NEVER ~≡0~(mod~8)

I f X , Y c a n t a k e a l l v a l u e s f r o m 0 t o 9 t h e r e w o u l d b e 100 s o l u t i o n s . If~X,~Y~can~ take~ all~ values~ from~ ~0~to~~9~there~ would ~ be~ 100~ solutions.

B u t f o r ( b , a ) = ( 1 , 2 ) , Y s h o u l d b e o n l y E V E N , w e g e t o n l y 50 S o l u t i o n s . But~for~(b,a)=(1,2),~~~Y~should~be~only~EVEN,~we~ get~ only~~50~~Solutions.

F o r ( b , a ) = ( 1 , 8 ) Y s h o u l d b e o n l y O D D . , w e g e t o n l y 50 S o l u t i o n s . For~(b,a)=(1,8)~Y~ ~should~be~ only~ODD. ,~we~ get~ only~~50~~Solutions. 50 + 50 = 100 ~\boxed{50+50=100}

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As a suggestion, I think you should clarify that the problem statment is atually, how many pairs of integers (a,b) such that b 2 − a 2 b^{2}-a^{2} is congruent to 2 mod 5 and congruent to 5 mod 8.

Evan Prout - 6 years, 9 months ago

b > a but u used (b,a) = (1,8)

Amar Mavi - 5 years, 3 months ago

What's the?I don't know the weather have being changed from when? However,This problem have no integer solution.Why? because for all integer x: x^2=0,1,4 mod 5 . Thus,There are no (b^2 -a^2)=2 mod5 that equivalent

Sereysopea Ung - 6 years, 9 months ago

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Please treat others with respect.

If b 2 ≡ 1 , a 2 ≡ 4 ( m o d 5 ) b^2 \equiv 1, a^2 \equiv 4 \pmod{5} , then we have b 2 − a 2 ≡ 2 ( m o d 5 ) b^2 - a^2 \equiv 2 \pmod{5} . That is how he concluded that

" ( b , a ) = ( 1 or 9 , 2 or 8 ) ( m o d 10 ) (b,a)=(1 \text{ or } 9, 2 \text{ or }8) \pmod{10} "

Calvin Lin Staff - 6 years, 9 months ago

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