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

• Born2Perform
In summary, the conversation discusses an exercise from a math olympiad and how to prove that the equation n=a²+b²-c² is true for any integer n, a, b, and c. The conversation also mentions the special cases of n=0 and the values of k for even and odd numbers. The main concept is that any integer can be expressed as a difference of two squares if the factors differ by an even number.
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?

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

e(ho0n3 said:
Certainly there is more to the problem. For example, pick n = 0 and a = b = c = 1. Then the equation is not true.
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

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 don't 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.

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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

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n must be even or odd or 0
Actually zero is an even number.

uart said:
Actually zero is an even number.

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,...

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Yeah it was ok to treat zero as a separate 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.

## 1. What does the equation N=a²+b²-c² represent?

The equation N=a²+b²-c² is known as the Pythagorean theorem, which states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

## 2. How does the Pythagorean theorem relate to the equation N=a²+b²-c²?

The Pythagorean theorem can be represented by the equation N=a²+b²-c², where N is the square of the length of the hypotenuse, and a and b are the lengths of the other two sides. This equation shows that the theorem holds true for any integer values of N, a, and b.

## 3. Can you provide an example to demonstrate the Pythagorean theorem using the equation N=a²+b²-c²?

For example, if we have a right triangle with a hypotenuse of length 5 and the other two sides have lengths of 3 and 4, we can use the equation N=a²+b²-c² to show that 5²=3²+4², proving the Pythagorean theorem.

## 4. How does the equation N=a²+b²-c² apply to non-right triangles?

The equation N=a²+b²-c² only applies to right triangles, as it is based on the Pythagorean theorem. For non-right triangles, other equations such as the Law of Cosines or the Law of Sines must be used to find the relationship between the sides and angles.

## 5. Why is it important to show that the equation N=a²+b²-c² is true for any integer values of n,a,b,c?

It is important to show the equation is true for any integer values of n,a,b,c because it is a fundamental concept in geometry and has many real-world applications, such as in architecture, engineering, and navigation. Understanding this equation and its validity for all integers allows us to solve a variety of problems and make accurate calculations in these fields.

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