# N=a²+b²-c², show that it's true for any integer n,a,b,c

1. Feb 17, 2007

### Born2Perform

n=a²+b²-c², show that it's true for any integer n,a,b,c

its an exercise of the math olympiad of my city... i know i should have posted at least a bit of my work, but i think there is a trick to solve this category of problems that i dnt know...where should i start???

2. Feb 17, 2007

### e(ho0n3

Certainly there is more to the problem. For example, pick n = 0 and a = b = c = 1. Then the equation is not true.

3. Feb 17, 2007

### Born2Perform

i think it was meant as "pick an integer n, now show it can always been written as a²+b²-c², where a,b,c are integers also" i just adapted the text becasue had problems with the translation

4. Feb 17, 2007

### uart

The following works for me. First let a=0 and consider what integers you can construct with just the difference of two sqaures $$b^2 - c^2$$. By considering the expansion you should easily be able to show that any odd number can be composed from the difference of the two squares alone.

Actually lots of even numbers can be made from the difference of the two sqaures as well, though not all. For example it's also easy to show that all integers that are a multiple of 4 can also be contructed, though you dont even need that result here. Once you've establish that b^2-c^2 can make any odd number then just leave a=0 for odd numbers and set a=1 to make all the evens.

Last edited: Feb 17, 2007
5. Feb 18, 2007

### ssd

n must be even or odd or 0.
The case of n=0 is trivial, a=b=c=0

n is even: n=2k, k=1,2,...
2k= 1^2 + k^2 - (k-1)^2

n is odd: n=2k-1, k=1,2,...
(2k-1)= 0^2 + k^2 - (k-1)^2

Last edited: Feb 18, 2007
6. Feb 18, 2007

### uart

Actually zero is an even number.

7. Feb 18, 2007

### ssd

Yes, you are right. But, here as the case of n=0 gives rise to a trivial solution of all 0's (a=b=c=0), I separately mentioned the case notwithstanding it was a repetition.
As a matter of fact, the values of 'k' I mentioned, should be k=...-2,-1,0,1,2,...

Last edited: Feb 18, 2007
8. Feb 18, 2007

### uart

Yeah it was ok to treat zero as a seperate case here, though not totally required. Many of the numbers have multiple ways of being expressed. For example zero could just as easily have been expressed as $$0 = 1^2 + 0^2 - 1^2$$, which is again just one plus an odd number and not a special case.

The result that I was hoping that the OP would come to by himself (by looking at the expansion of the difference of the two sqaures) was that any integer that can be expressed as a product of two factors can be expressed as a difference of two squares provided that those two factors differ by an even number. This of course means that any number that can be factored as odd times odd or even times even can be expressed as a difference of two squares. That in turn implies that all odd numbers and every second even number can be expressed in this way.