Prove that there is a positive integer that can be written as the sum of squares of positive integers in two different ways. For some reason this is not clicking in my head and I can not figure it out. Any help will be greatly appreciated and I would appreciate no straight forward answer since I am trying to understand.
FUTURECompEng 0 Newbie Poster
DavidB 44 Junior Poster
I think one way would be to just provide an example, wouldn't it? For example,
2^2 + 3^2 = 4 + 9 = 13
13 is a positive integer, so you have proved that there is a positive integer that can be expressed as the sum of the squares of positive integers.
In fact, the sum of squares of any integers (x, y, z, etc.) is always a positive integer:
x^2 + y^2 + z^2 + . . . is always going to be a positive integer.
The only way for it not to be a positive integer would be if x, y, z, etc. were non-integer numbers.
delta_frost 0 Newbie Poster
There exist many such examples.But the best example is the Charamichael number 1105.It can be expressed as the sum of squares of two integers in 4 different ways or so.
In general let the statement that the numbers be x,y and z which satisfy:
x^2 + y^2 = z
be true.
If x and y are co-prime,i.e their HCF is 1,one of them is divisible by 3,then the above property satisfies for certain numbers like
13^ + 6^2 = 205 = 3^2 + 14^2
15^2 + 8^2 = 289 = 17^2 + 0^2 (But here zero IS NOT a positive integer)
This is not a generalization;
1105 = 24^2 + 23^2 = 9^2 + 32^2 = 31^2 + 12^2 = 4^2 + 33^2
FUTURECompEng 0 Newbie Poster
Thanks to both of you I now understand, and from your explanations it seems that I was over complicating the question.
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