Given that x, y fulfilled the following conditions. I, x - y is a prime number. II, xy is a perfect square number. III, x » 2015. Calculate the smallest possible value of x
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@Kusma: maybe $x\ge 2015$? – Oleg567 Sep 07 '18 at 13:52
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I mean x is more than or equal to 2015 – NATHAN Sep 07 '18 at 14:01
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1To speed the search, I'd look for even squares $≥4030$. Even, since two numbers which multiply to give an odd square would differ by an even number (which pretty clearly can't be $2$ as you'd then need two perfect squares that differ by $2$). $≥4030$ since $x≥2015$ and it is pretty clear that $y≥2$. You can probably make an argument that $y≥3$ but I didn't sort that out. – lulu Sep 07 '18 at 14:03
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Is there any theorems which I can use to tackle the question in its best way? – NATHAN Sep 07 '18 at 14:09
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If $n^2=xy$ with $x-y$ prime, then the greatest common divisor of $x $ and $y$ must be a divisor of $x-y$, so must be $1$ or $x-y$. If the gcd is $1$, the fundamental theorem of arithmetic then tells you both $x$ and $y$ must be perfect squares. The difference between two perfect squares $x$ and $y$ is prime only for $x=(n+1)^2$, $y=n^2$ (see this question). Now all you have to do is find the smallest $n$ such that $2n+1$ is prime and $x=(n+1)^2\ge 2015$.
If the gcd is $x-y=p$ and is prime, then again by the fundamental theorem of arithmetic $\frac x p$ and $\frac y p$ must be perfect squares. So $p=x-y=A^2p-B^2p=p(A^2-B^2)$ for some positive integers $A$ and $B$, so $A^2-B^2=1$. This is impossible.
Kusma
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The gcd divides $x-y$, but so what? You have to exclude the possibility of it being equal to $x-y$. – Wojowu Sep 07 '18 at 15:00
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Indeed. We could still have $x=(x-y)^a A^2$,$y=(x-y)^b B^2$ with $a=1$ and $b$ odd. – Kusma Sep 07 '18 at 15:12
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$x-y=p$ so $x=y+p$ and $xy=y(y+p)=n^2$.
$p\ne 2$ because then $x=a+1$ and $y=a-1$ and $xy=a^2-1\ne n^2$. So $p$ is odd and $x,y$ have different parities.
If $x=bj, y=bk$ then $x-y=b(j-k)$ is composite; so $\gcd{x,y}=1$. Hence both $y$ and $(y+p)$ must be squares.
So look at the first square larger than $2015$. $45^2=2025$. Try $(y+p)=x=2025$ recalling that $y$ itself is a square. Try $y=44^2=1936$. As it turns out $2025-1936=89$ and $89$ is prime.
So $x=2025, y=1936$ gives the smallest possible value of $x$ that satisfies the conditions.
Keith Backman
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