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Find the pairs (Posted on 2020-10-13) Difficulty: 3 of 5
Determine the natural numbers m,n such as 85m-n4=4.

No Solution Yet Submitted by Danish Ahmed Khan    
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Hints/Tips maybe a full solution | Comment 5 of 22 |
This "proof" didn't turn out quite the way I expected, and leaves me convinced that the solution is unique but NOT that the proof is rigorous. I'd be curious to hear other opinions.


(building on Brian Smith's starting point)

take advantage of the factorization of n^4 + 4 = (n^2 + 2n + 2)(n^2 - 2n + 2) = ((n+1)^2 + 1)((n-1)^2 + 1)

These factors cannot both contain multiples of 5. If they did, their difference would also be a multiple of 5. Since their difference is 4n, n would have to be a multiple of 5 and n^4 + 4 could not be a multiple of 5 and could not be 85^m.

The same argument demonstrates that these factors cannot both contain multiples of 17.

So the two factors must be 5^m and 17^m for their product to be 85^m.

Since the 17^m > 5^m (m > 0), it must also be that:

(n+1)^2 + 1 = 17^m

and

(n-1)^2 + 1 = 5^m

the solution of m = 1,n = 3 is fairly clear at this point, but I'm more interested in whether there are others.


Consider the ratio of the two factors as a function of n. We can write that (with a bit of algebra) as:

(n^2 + 2n + 2) / (n^2 - 2n + 2) = (n^2 -2n + 2 + 4n) / (n^2 -2n + 2) = 1 + 4n / (n^2 -2n + 2)

This expression has a maximum at n = sqrt(2)

As n increases (past the maximum), this ratio monotonically decreases and approaches 1. But the ratio of 17^m / 5^m monotonically INCREASES in m, and is never less than its m=1 value of 3.4

Since increasing m requires increasing n AND increasing the ratio of the factors, but increasing n can only decrease that ratio, their point of equality at n = 3, m = 1 can be the only solution.


  Posted by Paul on 2020-10-14 13:12:45
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