Question

No of integers a b c for which a^2+b^2-8c=3

Answer 1

Answer:

all the integers are 0, 1, 2, 3, 4, 5, 6, 7 (modulo 8)

=> all the integer squares are 0, 1, 4 (modulo 8)

if there are three integers a, b, c s.t. a^2 + b^2 - 8c = 6

then, a^2 + b^2 = 6 (modulo 8)

but in modulo 8, a^2 and b^2 may only take the values 0, 1 & 4 so we cannot get the value 6 by adding two of them

Answer 2

Answer:

0

Step-by-step explanation:

Given : a2 + b2 - 8 c = 3 , So

a2 + b2 = 3 + 8 c --- ( 1 )

Case I : If a = Even number and b = Even number , So

Let a = 2 m and b = 2 n , Here ' m ' and ' n ' are positive integers .

Now we substitute these values in equation 1 and get :

( 2 m )²+ ( 2 n )² = 3 + 8 c

⇒4 m² + 4 n²= 3 + 8 c

⇒ 4 ( m² + n² ) = 3 + 8 c

Here , L.H.S. = 4 ( m² + n²) , That is multiple of " even number = 4 " , So

L.H.S. = Even number

And we know " Even number + Odd number = Odd number " and in R.H.S. " 8 c " is multiple of ' 8 ' so that is an even number and ' 3 ' is an odd number . So

R.H.S. = Odd number .

Therefore,

We get no solution when ' a ' and ' b ' both are even number .

Case II : If a = Odd number and b = Odd number , So

Let a = 2 m + 1 and b = 2 n + 1, Here ' m ' and ' n ' are positive integers .

Now we substitute these values in equation 1 and get :

( 2 m + 1 )² + ( 2 n + 1 )² = 3 + 8 c

⇒4 m² + 1 + 4 m + 4 n²+ 1 + 4 n = 3 + 8 c ( We know ( a + b )² = a² + b² + 2 a b )

⇒4 m² + 4 m + 4 n² + 4 n + 2 = 3 + 8 c

⇒ 2 ( m² + m + n² + n + 1) = 3 + 8 c

Here , L.H.S. = 2 ( m2 + m + n2 + n + 1) , That is multiple of " even number = 2 " , So

L.H.S. = Even number

And R.H.S. = Odd number , As we explained in case I .

R.H.S. = Odd number .

Therefore,

We get no solution when ' a ' and ' b ' both are odd number .

Case III : When any one of ' a ' is ' b ' is an odd number , We consider a = Even number and b = odd number , So

Let a = 2 m + 1 and b = 2 n + 1 , Here ' m ' and ' n ' are positive integers .

Now we substitute these values in equation 1 and get :

( 2 m )²+ ( 2 n + 1 )²= 3 + 8 c

⇒4 m² + 4 n² + 1 + 4 n = 3 + 8 c

⇒4 m²+ 4 n² + 4 n = 2 + 8 c

⇒ 4 ( m² + n² + n ) = 2 ( 1 + 4 c )

Here , L.H.S. = 4 ( m² + n² + n ) , That is divisible by " 4 " , But R.H.S. = 2 ( 1 + 4 c ) , That is not divisible by " 4 " ,

Therefore,

We get no solution when any one of ' a ' and ' b ' is an odd number .

From Case I , Case II and Case III we get that there is no value of a , b and c that can satisfy given equation . So

Option ( C ) ( Ans)

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