How to (quickly) prove that $24p+17$ is not a square number Computer says $24p+17$ is not square number for $p<10^7$ so I guess it's not. I know that squares of odd numbers are all $8p+1$ but $24p+17$ passes the test
And how to solve problems like this in general? Thx in advance
 A: Reduce the equation modulo three. Since 3 divides 24 and since $17 = 5\times 3 + 2$ we see that $24p+17 \equiv 2\bmod 3$ for all $p$. Now notice that $0^2 = 0 \equiv 0 \bmod 3$, $1^2 =1 \equiv \bmod 3$ and $2^2 = 4 \equiv 1 \bmod 3$. So no square number is congruent to 2 modulo 3, meaning $24p+17$ is not square.
In general, I looked at the factors of 24. They are 1, 2, 3, 4, 6, 8, 12 and 24. The aim now is to reduce $24p+17$ modulo one of these factors. You then need to find a factor that when you reduce 17 modulo that factor, you get a number which is not given by squaring any of the members of $\mathbb{Z}_n$.
For example, let's try $n=6$. $24p+17 \equiv 5 \bmod 6$. Next $0^2 \equiv 0 \bmod 6$, $1^2 \equiv 1 \bmod 6$, $2^2 \equiv 4 \bmod 6$, $3^2 \equiv 3 \bmod 6$, $4^2 \equiv 4 \bmod 6$ and $5^2 \equiv 1 \bmod 6$. Again, no square is congruent to 5 modulo 6, while $24p+17$ is congruent to 5 modulo 6. Again, this proves that $24p+17$ is not a square.
A: $24$ is the largest number $m$ with the property that $x^2=1$ mod $m$ for all $x$ coprime to $m$. 
This follows from the Chinese remainder theorem. It implies that among the residue classes coprime to $24$, only the numbers of the form $24p+1$ contains perfect squares.
A: The problem is equivalent to “Prove that in Base 24, no square number ends in the digit H (i.e., seventeen).”
Reading down the diagonal of the quadrovigesimal multiplication table gives the square numbers as: 1, 4, 9, G, 11, 1C, 21, 2G, 39, 44, 51, 60, 71, 84, 99, AG, C1, DC, F1, GG, I9,
 K4, M1, 100.
Thus, in quadrovigesimal, all square numbers end in one of the digits {0, 1, 4, 9, C, G}, which does not include H.
A: The number is $2 \pmod 3$. No such number is perfect square.
If you are not familiar with the modular notation, any perfect square is either of the form $3k$ or $3k+1$. Your number is $3k+2$.
A: Any number not divisible by $6,$ can be expressed as $6p\pm1, 6p\pm 2,6p\pm 3$ where $p$ is any integer
If $p$ is an odd natural number, as $24p+17$ is odd and is not divisible by $3,$ its square root(if any) must of the form $6p\pm1$  as prime $q$ divides integer $n\iff q$ divides $n^2$ 
$$\text{Now, }(6p\pm1)^2=36p^2\pm12p+1=24p^2+24\frac{p(p\pm1)}2+1\equiv1\pmod {24}$$
Also (it's not required here), 
$$(6p\pm2)^2=36p^2\pm24p+4=12p^2\pm24p+4=12(p^2-1)\pm24p+16\equiv16\pmod {24}\text{ as } 2  (\text{ in fact } 8 )\text{ divides } (p^2-1) \text{ for odd  }p$$
$$(6p\pm3)^2=36p^2\pm36p+9=72\frac{p(p\pm1)}2+9\equiv9\pmod {24}$$
