Here x and y range over the nonnegative integers. I was wondering how to show that if n is a odd integer which has a unique representation in that form, then n is prime.
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HINT $\ $ Nonuniquness is an immediate consequence of the following composition law $\rm\qquad\qquad\ (a^2-b^2)\ (A^2-B^2)\ =\ (a\:A+b\:B)^2-(a\:B+A\:b)^2$ $\rm\qquad\qquad\ \phantom{(a^2-b^2)\ (A^2-B^2)}\ =\ (a\:A-b\:B)^2-(a\:B-A\:b)^2$ E.g. composing $\rm\ 7 = 4^2 - 3^2\ $ with $\ 11 = 6^2 - 5^2\ $ yields for $\rm\: 7\cdot 11\:$ the following $2$ rep's $\rm\qquad\qquad\ (4^2-3^2)\ (6^2-5^2)\ =\ (4\cdot 6+3\cdot 5)^2-(4\cdot 5+6\cdot 3)^2\ =\ 39^2 - 38^2$ $\rm\qquad\qquad\ \phantom{(4^2-3^2)\ (6^2-5^2)}\ =\ (4\cdot 6-3\cdot 5)^2-(4\cdot 5-6\cdot 3)^2\ =\ 9^2 - 2^2$ |
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Hint: Factor $x^2-y^2$ as the product of two polynomials. |
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Hint: Let $n$ be an odd composite (positive) integer. Then there exist (odd) integers $a \gt 1$, $b \gt 1$ such that $n=ab$. Use this factorization to produce a solution of the equation $x^2-y^2=n$ in non-negative integers other than $x=\left(\frac{n+1}{2}\right)^2$, $\:\;y=\left(\frac{n-1}{2}\right)^2$. Added: Without loss of generality we may assume that $a \ge b$. Then we can take $$x=\left(\frac{a+b}{2}\right)^2, \qquad y=\left(\frac{a-b}{2}\right)^2.$$ |
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