Does it possible to show that the Diophantine equation $X^2-Y^2=N$ (N - odd)has no non-trivial solution?
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$\begingroup$ No. $(t+1)^2-t^2=2t+1$ is odd. Did you mean maybe $N^2$ instead of $N$? $\endgroup$– coffeemathCommented Oct 15, 2015 at 15:52
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$\begingroup$ For more cases, if $k$ is odd then $(t+k)^2-t^2=2tk+k^2$ is also odd. By the way what does "trivial" mean, does it mean that say $Y=0$? $\endgroup$– coffeemathCommented Oct 15, 2015 at 16:00
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1$\begingroup$ By trivial I mean &X=(N+1)/2; Y=(N-1)/2& $\endgroup$– Boris SklyarCommented Oct 15, 2015 at 16:05
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$\begingroup$ Boris. OK, see answer below, it seems it may have nontrivial solutions. $\endgroup$– coffeemathCommented Oct 15, 2015 at 16:13
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2 Answers
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Suppose $N=CD$ where $C,D$ are odd and greater than $1$ and $C<D.$
Then setting $X-Y=C,\ X+Y=D$ gives $X=(D+C)/2,\ Y=(D-C)/2.$ Since these are not $(N \pm 1)/2$ the solution is not "trivial" in the sense of your last comment.
answered Oct 15, 2015 at 16:11
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$\begingroup$ What if N - is a prime number? $\endgroup$ Commented Oct 15, 2015 at 16:15
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$\begingroup$ @Boris Your question didn't say $N$ was a prime. But if it is an odd prime $p,$ then given $X^2-Y^2=(X-Y)(X+Y)=p$ the only possibility is $X-Y=1,X+Y=p$ which leads to the "trivial" situation you formulate in your comment. $\endgroup$ Commented Oct 15, 2015 at 16:24
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$\begingroup$ My aim is to use the following statement: Odd number N is a prime if and only if the Diophantine equation $X^2−Y^2=N$ has no non-trivial solution, for checking primality of N $\endgroup$ Commented Oct 15, 2015 at 16:33
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$\begingroup$ @Boris Then your statement is valid, as shown by the example in my answer a non-prime odd has nontrivial solutions, and if $N$ is prime there is only the trivial solution as just noted. So if that's your aim it is proved. $\endgroup$ Commented Oct 15, 2015 at 16:35
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$\begingroup$ But I need to show that for given N there is no non-trivial solution (N - prime) or to show that for given N there is non-trivial solution (N - composite). Does such algorithm exist? $\endgroup$ Commented Oct 15, 2015 at 16:43
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Basically, the equation $$x^2-y^2=N$$ is such that the set of integer solutions $(x,y)\in {\mathbb Z}^2$ has a bijection (given in older posts) with the set of integer solutions $(a,b)$ such that $a \pm b\equiv N (\mbox{mod }2)$ and $$ab=N.$$
Thus, a natural number $N>0$ has no nontrivial solutions if and only if $N$ is a prime number (odd case) or a number $\equiv 2 (\mbox{mod }4)$.
I am merely rephrasing what's been done above, i.e. clearly $a=x-y$, and $b=x+y$.
