Since everything is squared and since 1+4=5, it looks like two instances of a difference of squares. So this equals i^2-k^2+j^2-4k^2 = (i-k)(i+k) + (j-2k)(j+2k).
The easiest way to continue is for i-k and j-2k to 1, 0, or 1, so they drop out and simplify the equation in only terms of k.
i-k=1, i-k=0, i-k=-1
6k+2, 4k+1, 2k+2
2k+1, 0, -2k+1
-2k+2,-4k+1, -6k+2
Looking at these, 2k+2 and 2k+1 account for all cases with an integer k. So if n=2k+2, i-k=-1 (so i=k-1) and j-2k=1 (so j=2k+1) Solving for k gives (n/2)-1=k and substituting that into these gives i=(n/2)-2 and j=n-1.
Also, if n=2k+1, i-k=1 (so i=k+1) and j-2k=0 (so j=2k), and this means (n-1)/2=k, so i=(n+1)/2 and j=n-1.
For example, if n=14, k=6, i=5, j=13. This makes sense since 13^2+5^2-5(6^2)=14. For n=17, k=8, i=9, j=16, so 16^2+9^2-5(8^2)=17.
Substituting into the original equation for 2k+2 gives n^2/4 - 2n + 4 + n^2 - 2n + 1 - 5n^2/4 + 5n - 5, which simplifies to just n. Also, substituting into the original equation for 2k+1 gives n^2/4 + n/2 + 1/4 + n^2 - 2n + 1 - 5/4 n^2 + 5n/2 - 5/4 which equals n. This verifies that these two sets of equations for i, j, and k work to simplify to n.
I would bet that the other possiblities for n (n=6k+2 and n=4k+1) would work too for their own values of i and j, but most integers can't be expressed that way, and it is easier to use the previous two equations. Also, 2k+1 still works for even values of n, but it would result in i, j and k not being integers.
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Posted by Gamer
on 2005-11-24 22:20:05 |
Solution
