Similar to USA TST 2008. From $a)$, we notice that $2017 = 45^2 - 2^3$; using this on $b)$ we have that $y^2 + 45^2 = (x+2)(x^2 - 2x + 4)$. From the original equation we notice that $x \equiv 1 \pmod 4$ and hence $x + 2 \equiv 3 \pmod 4$. Hence there is a prime factor $p \equiv 3 \pmod 4$ s.t. $p | x+2$ and $p | y^2 + 45^2$. Then $p | 45$ and $p | y$. But the only prime factor $\equiv 3 \pmod 4$ from $45$ is $3$. Therefore, $p = 3; 3 | y$, a contradiction. $\blacksquare.$

hydo2332 wrote: Similar to USA TST 2008. From $a)$, we notice that $2017 = 45^2 - 2^3$; using this on $b)$ we have that $y^2 + 45^2 = (x+2)(x^2 - 2x + 4)$. From the original equation we notice that $x \equiv 1 \pmod 4$ and hence $x + 2 \equiv 3 \pmod 4$. Hence there is a prime factor $p \equiv 3 \pmod 4$ s.t. $p | x+2$ and $p | y^2 + 45^2$. Then $p | 45$ and $p | y$. But the only prime factor $\equiv 3 \pmod 4$ from $45$ is $3$. Therefore, $p = 3; 3 | y$, a contradiction. $\blacksquare.$ How did you go from line 9 to line 10? How is it necessarily true?

B) Mordells Equation $y^2 = x^3-2017$ $y^2+2017 = x^3$ $(y+\sqrt{-2017})(y-\sqrt{-2017}) = x^3$ $k = a+b\sqrt{-2017}$ $c = c+d\sqrt{-2017}$ $y+\sqrt{-2017} = k^3$ $y-\sqrt{-2017} = c^3$ $y+\sqrt{-2017} = (a+b\sqrt{-2017})^3$ $y= a^3-6051ab^2$ $\sqrt{-2017} = \sqrt{-2017}(b)(3a^2-2017b^2)$ $1 = (3a^2-2017b^2)(b)$ Case 1 $b = 1$ $2018 =3a^2$ $+-\sqrt{\frac{2018}{3}}=a$ $(-\sqrt{\frac{2018}{3}},1),(+\sqrt{\frac{2018}{3}},1)$ Case 2 $b=-1$ $1=(3a^2-2017b^2)(-1)$ $-1 = 3a^2-2017$ $2016 = 3a^2$ $+-\sqrt{672}=a$ $(\sqrt{672},-1),(-\sqrt{672},-1)$ $y = a^3-6051(ab^2)$ $y = a(a^2-6051b^2)$ we can easily see that there is no y that is integer

(a) Take $(x,y)=(2,45).$ (b) Suppose FTSOC that $3\nmid y$ and rewrite the given equation as \[y^2+45^2=(x+2)(x^2-2x+4).\]By Fermat's Christmas Theorem, any prime $p\equiv 3\pmod{4}$ that divides $y^2+45^2$ must divide $\gcd(y,45)=\gcd(y,5).$ Hence, there exists no such primes $p$ so $x+2\equiv 0,1,2\pmod{4}.$ If $x\equiv -2,0\pmod{4},$ then $x^2-2x+4\equiv 0\pmod{4}$ so we need $y^2+45^2\equiv 0\pmod{4},$ which is absurd. If $x\equiv -1\pmod{4},$ then $x^2-2x+4\equiv 3\pmod{4},$ a contradiction. $\square$

For part (a), use $(2, 45)$. For part (b), rewrite the equation as $$y^2 + 2025 = x^3+8.$$Notice that evidently $x$ must be odd, otherwise the LHS is 2 mod 4. Assume for the sake of contradiction that $3 \nmid y$, thus $\gcd(y, 2025) \mid 25$. By Fermat Christmas, every divisor of $x^3+8$ must be 1 mod 4 (including the divisor $5$ in the GCD, if it exists.) But this implies that $x+2 \equiv 1 \pmod 4$ and $x^3+8\equiv 1 \pmod 4$ simultaneously, th first of which yields $x \equiv 3 \pmod 4$ and the second of which yields $x \equiv 1 \pmod 4$. This is obviously impossible.

ericxyzhu wrote: hydo2332 wrote: Similar to USA TST 2008. From $a)$, we notice that $2017 = 45^2 - 2^3$; using this on $b)$ we have that $y^2 + 45^2 = (x+2)(x^2 - 2x + 4)$. From the original equation we notice that $x \equiv 1 \pmod 4$ and hence $x + 2 \equiv 3 \pmod 4$. Hence there is a prime factor $p \equiv 3 \pmod 4$ s.t. $p | x+2$ and $p | y^2 + 45^2$. Then $p | 45$ and $p | y$. But the only prime factor $\equiv 3 \pmod 4$ from $45$ is $3$. Therefore, $p = 3; 3 | y$, a contradiction. $\blacksquare.$ How did you go from line 9 to line 10? How is it necessarily true? If $p$ is prime and of the form $4k+3$ it divides $m^2+n^2$ if and only if $p|m$ and $p|n$