As the prime numbers have to be expressed as the difference of squares of two CONSECUTIVE integers, the representation is unique for each prime number. Prove also that this presentation is unique: THE REPRESENTATION IS UNIQUE ONLY FOR PRIME NUMBERS.Prove that every odd prime number can be written as a difference of two squares: All numbers can be, due to the $(a+b)(a-b)$ representation.BUT, for any other even or odd, non-prime numbers, this representation WILL NOT be unique, as they will have more than 2 factors. So, for primes, we have found them to be consecutive using earlier logic. Now, here, $a+b$ and $a-b$ are two factors of the number. So, for any number, if it can be written as a difference of squares, it can also be written as $(a+b)(a-b)$ Note that, in the beginning, we have noticed that: As all primes are odd, they can also be represented in that way. So, the difference of squares of two consecutive positive integers covers ALL the odd numbers. So, the difference of squares of n and $n+1$, and $n+1$ and $n+2$ gives us $2n+1$ and $2n+3$ respectively, which are two consecutive odd numbers. Now, the positive integer just after n+1 is n+2. Therefore, we have proved that every prime number can be expressed as $a^2-b^2$, where a and b are positive integers (and also consecutive positive integers).Īnother way of proving this in a much simpler and algebraic way (which is easy to do once you know the fact that a and b in the above proof are consecutive positive integers) is this: So, we have also proved that every odd prime number can be expressed as the sum of two consecutive positive integers. Hence, every odd number can be expressed as the sum of two consecutive positive integers. We know that every odd number can be expressed as 2k+1, where k is a positive integer. Now, we only have to deal with odd primes (all primes except 2). Therefore, if we can prove that every prime number can be expressed as the sum of two consecutive positive integers, we will also prove that every prime number can be expressed as the difference of two squares. So, a and b are two consecutive positive integers! Now, we know that: Now, p only has 1 and p as factors, therefore, either $(a+b) = p$ and $(a-b) = 1$, or $(a-b) = p$ and $(a+b) = 1$.īut, as a and b are positive integers, $(a+b) > (a-b)$. So, we have found two factors of a number, p, which is prime. I will prove that p can be expressed as $a^2-b^2$ and also show that we can represent p in that way for a certain relationship between a and b. We have to prove that every prime number can be expressed as $a^2-b^2$, where a and b are positive integers. I have found some interesting solutions and proofs by myself that I am sharing. I have come across this problem in a book that I was reading.
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