question about numbers?

Question

Is it possible for 8*n-3 to be a perfect square?

For all you people who say yes, give me a proof! Any idiot can say "oh, yeah, that's a perfect square." If it is, give me a value for n. Otherwise your answer is completely useless!

u4--very nice, thank you.

Answer 1

No, assuming that you mean n must be a positive integer.(Otherwise, n could be 19/8, etc.)

If 8n-3 is a perfect square for some positive integer n, then 3+z^2 must be divisible by 8 for some positive integer z.

Now note that the remainder of 3+z^2 divided by 8 is equal to the remainder of 3+(z+4)^2=3+z^2+8z+16 divided by 8, since 8z+16 is always divisible by 8, and thus will not contribute to a remainder.

So this means the remainder of 3+z^2 repeats every for every four values of z in sequence!

This is easiest illustrated in a table:

z __z^2+3__ remainder
1 ___4___ 4
2 ___7___ 7
3 ___12___ 4
4 ___19___ 3
5 ___28___ 4
6 ___39___ 7
7 ___52___ 4
8 ___67___ 3
9 ___84___ 4
....

(___ used to space out numbers since yahoo kills extra spaces.)

You can see the pattern for the remainder is 4,7,4,3.

Since none of the first 4 remainders are zero, it follows by induction that the remainder will never be zero, so 3+z^2 will never be divisible by 8, and thus, 8n-3 will never be a perfect square.

Answer 2

others have provided ans and let me provide a simpiler one

a number is of the form 2k or 2k+1
8n-3 is odd so we can take 2k+1

now (2k+1)^2 + 3 = 4k^2+ 4k + 1+ 3
= 4k(k+1) + 4

1st term is divisible by 8 as either n or n+1 is odd.

so above number divided by 8 leaves a remainder 4

so (2k+1)^3+ 3 is not divisible by 8
so 8n-3 cannot be a perfect square

Answer 3

yes it's a perfect square

Answer 4

No because the answer will always be prime or not a square.

Answer 5

8n-3=k^2, k=natural number
it has not any answer in natural numbers.

The residue of k on 8 can be 0,1,2,3,4,5,6,7.
The residue of k^2 on 8 can be 0,1,4,1,0,1,4,1.
The residue of k^2 + 3 on 8 can be just 3,4,7.
it means that it has not answer, it has always residue.

Answer 6

no it's not possible
(8*1-3=5
8*2-3=13
8*3-3=21
and so on)
it will always come to be a prime no.

Answer 7

Only if n is not constrained to integer values.