If $x$ is odd then $y$ is even, thus $y^3\equiv 0\pmod{4}$ whereas $x^2+5\equiv 2\pmod{4}$, a contradiction. Hence, $x$ is even. Consequently, $y\equiv 1\pmod{4}$. Now, $x^2+4 = (y-1)(y^2+y+1)$. Check that $y^2+y+1\equiv 3\pmod{4}$, thus there is a prime $q>2$, congruent to $3$ modulo $4$ so that $q\mid y^2+y+1$. Hence, $q\mid x^2+2^2$. Setting $a\equiv x\cdot 2^{-1}\pmod{q}$ (check that $q>2$ implies $2^{-1}$ modulo $q$ exists), we deduce $a^2+1\equiv 0\pmod{q}$. This is impossible as $q\equiv 3\pmod{4}$.

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parmenides51 wrote: Show that there are no integers $x$ and $y$ satisfying $x^2 + 5 = y^3$. Daniel Harrer Use mod 4

hi can any one plz help me solve this find all integers a,b>=0 such as 3a²=5^b +2