Logarithm question, finding all possible pairs of integers

[trot]

Homework Statement

Find all possible pairs of integers a and n such that:

log(1/n)(√(a+√(15)) - √(a -√(15)))=-1/2

(that's log to the base (1/n))

The Attempt at a Solution

(1/n)^-1/2 = (√(a+√(15)) - √(a -√(15))
∴ n^4 = (a+√(15) - (a -√(15) - 2√((a+√15)(a -√(15))
∴ n^4 = =2√(15) - 2√((a+√15)(a -√(15))
eventually simplifying to:
n^(16)-√(15)n^4 =4a^2

dont know how to solve, probably made mistake

question is from core 3 edexcel and is worth 13 marks

 

[micromass]

(1/n)^-1/2 = (√(a+√(15)) - √(a -√(15))
∴ n^4 = (a+√(15) - (a -√(15) - 2√((a+√15)(a -√(15))

I really don't understand this step.

Also, it would help us a lot if you would type up your posts in LaTeX. Here is a basic guide: https://www.physicsforums.com/showpost.php?p=3977517&postcount=3

 

[symbolipoint]

Typing in all the steps into the Compose form would be a mess; but one of my partial results seems to give the equation,
(n-2a)/(-2) = sqrt(a²-15)

 

[micromass]

Typing in all the steps into the Compose form would be a mess; but one of my partial results seems to give the equation,
(n-2a)/(-2) = sqrt(a²-15)

Yeah, I get the same thing.

 

[symbolipoint]

Further steps give me n²-4an+60=0, from which solving for a,
a=(n/4)+(15/n).

That could let us find possible pairs of INTEGERS for a and n.

 

[micromass]

Further steps give me n²-4an+60=0, from which solving for a,
a=(n/4)+(15/n).

That could let us find possible pairs of INTEGERS for a and n.

OK, but maybe we should let the OP solve it?

 

[symbolipoint]

OK, but maybe we should let the OP solve it?

Knowing no clever way to solve specifically for integer solutions, would a BASIC FOR loop be acceptable? I would run n from about 0.100 to 50, incrementing by 0.100 for each step. a would be calculated in each run through the loop.

( I know no clever, fancy way to find the integer solutions for this rational equation but I believe a BASIC program can expose some integer number pairs for n and a ).