If $n$ is known,and $x,y,n$ belong to $\mathbb{N}^+$. What is $x$ and $y$? I know there exists a answer, for example,when $n=111$, $x=6161$, $y=6160$, but I do not know if the answer is unique.
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Hint. Look for solutions when $n=2,3,4$ and $8$. – Ethan Bolker Feb 23 '16 at 16:18
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@EthanBolker Why those numbers specifically? – Airdish Feb 23 '16 at 16:23
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Notice that $x^2-y^2=(x-y)(x+y)$. Notice also that the solution is unique if $2n-1$ is a prime number. – Giovanni Resta Feb 23 '16 at 16:26
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The answer is not generally unique: $41^2 - 40^2 = 15^2-12^2 =9^2$. – Andrew Dudzik Feb 23 '16 at 16:28
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@TheOddbodNumber Several of the other comments hint at where nonuniqueness happens. $15^2$ is interesting because $15$ is the product of two odd primes, so gives lots of nonuniqueness. That's why I suggested $n=8$. – Ethan Bolker Feb 23 '16 at 16:44
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There are numerous solutions where at least many (infinity) may be calculated based on the answer. – Moti Feb 24 '16 at 05:28
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Liam, I posed an answer for you! – Moti Feb 25 '16 at 19:31
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@Moti Thank you for your help. I cannot vote because of my limited reputation. – liam xu Feb 25 '16 at 20:10
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My take:
Assuming $a, b$ whole numbers, let replace $x=a^2+b^2$ and $y=2ab$ and we get $2n-1=a^2-b^2$, (or $2n-1=b^2-a^2$)
The only limitation on $a, b$ is that one is odd and the other even, and $a>b$
Example: a=4, b=3 results x=25, y=24, and n=4
Moti
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