Prove that $5a^2 \equiv k \bmod 12$ where $k \in \{0,5,8,9\}$.

Question

Prove that $5a^2 \equiv k \bmod {12}$ where $k \in \{0,5,8,9\}$.
Hence show that the equation $24x^7 + 5y^2 = 15$ has no integer solutions.

I think I need to evaluate each case with the $k$ values first but not sure how...

Answer 1

$a\equiv0,\pm1,\pm2,\pm3,\pm4,\pm5,6$

$$\implies a^2\equiv0,1,4,9$$

$$\implies k\equiv5a^2\equiv0,5,,20\equiv8,45\equiv9$$

$$24x^2+5y^2=15\implies5y^2\equiv15\pmod{12}\iff y^2\equiv3$$

Answer 2

Consider the following cases:

• $a\equiv 0\pmod{12} \implies 5a^2\equiv5\cdot 0^2\equiv 0\equiv0\pmod{12}$
• $a\equiv 1\pmod{12} \implies 5a^2\equiv5\cdot 1^2\equiv 5\equiv5\pmod{12}$
• $a\equiv 2\pmod{12} \implies 5a^2\equiv5\cdot 2^2\equiv 20\equiv8\pmod{12}$
• $a\equiv 3\pmod{12} \implies 5a^2\equiv5\cdot 3^2\equiv 45\equiv9\pmod{12}$
• $a\equiv 4\pmod{12} \implies 5a^2\equiv5\cdot 4^2\equiv 80\equiv8\pmod{12}$
• $a\equiv 5\pmod{12} \implies 5a^2\equiv5\cdot 5^2\equiv125\equiv5\pmod{12}$
• $a\equiv 6\pmod{12} \implies 5a^2\equiv5\cdot 6^2\equiv180\equiv0\pmod{12}$
• $a\equiv 7\pmod{12} \implies 5a^2\equiv5\cdot 7^2\equiv245\equiv5\pmod{12}$
• $a\equiv 8\pmod{12} \implies 5a^2\equiv5\cdot 8^2\equiv320\equiv8\pmod{12}$
• $a\equiv 9\pmod{12} \implies 5a^2\equiv5\cdot 9^2\equiv405\equiv9\pmod{12}$
• $a\equiv10\pmod{12} \implies 5a^2\equiv5\cdot10^2\equiv500\equiv8\pmod{12}$
• $a\equiv11\pmod{12} \implies 5a^2\equiv5\cdot11^2\equiv605\equiv5\pmod{12}$