8
$\begingroup$

(Please read "Edit"s and see this.)

How could I prove that : $\text{If} \space m^2=a^3-b^3\text{ where}\space m,a,b\in\mathbb{N} \rightarrow \exists c,d \in\mathbb{N}\space \text{ such that}\space m=c^2+d^2 $ thanks for helping

Edit: I told the person who gave me this question it's wrong, and he corrected it like this: $\text{If} \space m^2=(a+1)^3-a^3\text{ where}\space m,a\in\mathbb{N} \rightarrow \exists c,d \in\mathbb{N}\space \text{ such that}\space m=c^2+d^2 $ It's such an easy question and I already know the answer.

Edit2:I though I know this question answer but after thinking I can't solve this, could any one help me to figure out how to solve this?(I hope it wasn't wrong like previous question, but if you think it's wrong please let me know,I need to solve this question for exam I wanna take from my students.)

Edit 3: the second question wasn't wrong and has been answered at this link.

  • 0
    @JonasMeyer I dont know some one ask this from me2012-01-16

1 Answers 1

24

You cannot prove it because it is false.

Let $a=90$, $b=54$. Then $a^3-b^3=571536=2^4\cdot3^6\cdot7^2$ and hence $m=2^2\cdot 3^3\cdot 7.$ This cannot be written as the sum of two squares since it has prime factors congruent to $3$ modulo $4$ which appear to an odd power. (Violating Fermat's condition)

Also see: http://mathworld.wolfram.com/SumofSquaresFunction.html

Edit: The smallest example occurs when we take $a=10$, $b=6$, as then $a^3-b^3=784=(28)^2$ so that $m=2^2\cdot 7$. This cannot be written as the sum of two squares since we cannot write $7$ as the sum of two squares.

  • 0
    @AliAmiri: Yes, I accidently put the square2012-01-19