Number Thry: integer soln of [tex]\sqrt{2n^2\pm5}[/tex] & [tex]\sqrt{2n^2\pm7}[/tex]

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math_fan

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Number Thry: integer soln of [tex]\sqrt{2n^2\pm5}[/tex] & [tex]\sqrt{2n^2\pm7}[/tex]

\(\displaystyle \sqrt{2n^2\pm7}\) appears to infinitely many integer solutions (n = 1, 2, 5, 12, 29, ....) whereas

\(\displaystyle \sqrt{2n^2\pm5}\) appears to have none.

Is there someway to determine if \(\displaystyle \sqrt{2n^2\pm\ x}\) has integer solutions for a given x without trial and error?


Many thanks
 
math_fan said:
\(\displaystyle \sqrt{2n^2\pm7}\) appears to infinitely many integer solutions (n = 1, 2, 5, 12, 29, ....) whereas

\(\displaystyle \sqrt{2n^2\pm5}\) appears to have none.

Is there someway to determine if \(\displaystyle \sqrt{2n^2\pm\ x}\) has integer solutions for a given x without trial and error?


Many thanks
Click to expand...

I don't think it can be determined explicitly.
Certain numbers can be discounted as follows...

\(\displaystyle 2n^2\equiv 0\ or\ 2\ MOD\ 8\)

so require \(\displaystyle x\equiv 0,\ 1,\ 2,\ 4,\ 6\ or\ 7\ MOD\ 8\)

so any \(\displaystyle x\equiv 3\ MOD\ 8\ or\ x\equiv 5\ MOD\ 8\) will definitely have no solutions.

I think it's trial and error for the others though.

Anyone have anything better? (hopefully)
 
Last edited:
Denis said:
Where/how did you get those solutions?
I get n = 1,2,3,4,8,9,19,22,46,53 ; no more under 100.
And one solution for each, either the + or the -
Click to expand...
\(\displaystyle \sqrt{2 * 1^2 - 7} = \sqrt{2 * 1 - 7} = \sqrt{2 - 7} = \sqrt{- 5} \not\in \mathbb Z.\)

Corner time.
 
JeffM said:
\(\displaystyle \sqrt{2 * 1^2 - 7} = \sqrt{2 * 1 - 7} = \sqrt{2 - 7} = \sqrt{- 5} \not\in \mathbb Z.\)

Corner time.
Click to expand...

The other solution exists:

\(\displaystyle \sqrt{2 * 1^2 \pm 7} = \sqrt{2 * 1 + 7} = \sqrt{2 + 7} = 3\)

Corner time.
 
Denis said:
Where/how did you get those solutions?
I get n = 1,2,3,4,8,9,19,22,46,53 ; no more under 100.
And one solution for each, either the + or the -
Click to expand...

Quite right Denis.

I was merely showing there were integer solutions for \(\displaystyle \sqrt{2n^2\pm 7}\)

The equation I was working with was \(\displaystyle (m^2-2mn-n^2)^2=49\)

This solves for \(\displaystyle m=n\pm\sqrt{2n^2\pm 7}\)

The other values you have included are the complimentary \(\displaystyle m\)s for each given n
 
The types of example that are beating me are where the result is a 'possible' square. i.e

\(\displaystyle if\ \ 2n^2\equiv 2\ MOD\ 8\ \ \ and\ \ \ x\equiv 6\ MOD\ 8\ \ \ then\ \ \ 2n^2-x\equiv 4\ MOD\ 8\ \ \ and\ \ \ 2n^2+x\equiv 0\ MOD\ 8\)

This means this 'could' produce squares for both the + and - cases, but never appears to.
 
solved

math_fan said:
The types of example that are beating me are where the result is a 'possible' square. i.e

\(\displaystyle if\ \ 2n^2\equiv 2\ MOD\ 8\ \ \ and\ \ \ x\equiv 6\ MOD\ 8\ \ \ then\ \ \ 2n^2-x\equiv 4\ MOD\ 8\ \ \ and\ \ \ 2n^2+x\equiv 0\ MOD\ 8\)

This means this 'could' produce squares for both the + and - cases, but never appears to.
Click to expand...

Sorry guys, must head for corner myself :(

I have found valid results for all cases where the solutions can be integers. I misread my table when writing the above. Both cases can be solved, I was looking at

\(\displaystyle 2n^2\pm x\equiv 6\ MOD\ 8\)

which obviously cannot be an integer square.

 
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