Takes me back to 1980's!!
How many integer solutions exists for
x 2 + y 2 + z 2 = 1 9 8 0 ?
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4 solutions
thanx for that piece of info.!!
what are a and b?
a and b are any number possible ......with the help of which we can equate left and right side ..if there is any value of a and b satisfy it..thn there is no solution ..like here a =1 and b =61
Please can you recommend some good books.
x 2 + 0 x + ( y 2 + z 2 − 1 9 8 0 ) = 0
For solutions in x, b 2 < = 4 a c so:
y 2 + z 2 − 1 9 8 0 < = 0
y 2 + z 2 < = 1 9 8 0
It follows that when solving for y and z,
x 2 + y 2 < = 1 9 8 0
x 2 + z 2 < = 1 9 8 0
Adding these together,
2 x 2 + 2 y 2 + 2 z 2 < = 1 9 8 0
x 2 + y 2 + z 2 < = 9 9 0 which isn't consistent with original equation. So no solutions.
Mistake on RHS after adding 3 inequalities
why don't you ask Legendre..
I think the "only if" part is simple to prove, and this is all that is needed to answer the question. The "only if" part says a number, n , cannot be written as a sum of three integers if it is 4 a times a number that is 7 mod 8.
First if n does not have 4 a as a factor, this requires showing that the sum of three squares, mod 8, cannot be 7. As a square, mod 8, is either 0,1 or 4, this is immediate.
If n has a factor of 4 a , you can first divide both sides by 4 a . To get
( x / 2 a ) 2 + ( y / 2 a ) 2 + ( z / 2 a ) 2 = n / 4 a
and for this to have a solution each of ( x / 2 a ) , ( y / 2 a ) and ( z / 2 a ) will need to be integers. But there do not exist three integers whose square is equal to n / 4 a , as that number is 7 mod 8.
http://en.wikipedia.org/wiki/Legendre's three-square theorem
According to Legendre 's three square theorem
"any number "n" can be expressed as the sum of three squares of integers only and only if
n≠ 4 a (8b+7)
We have,
4.(8(61)+7)=1980
Thus it has no solutions when x, y and z are integers.