# Tag Info

### Proof that square of an $\sqrt{2 - 4q}$ is not an integer

Hint : If $n$ is some integer, then $n=0 \mod{4} \implies n^2=0 \mod{4}$ $n=1 \mod{4} \implies n^2=1 \mod{4}$ $n=2 \mod{4} \implies n^2=0 \mod{4}$ $n=3 \mod{4} \implies n^2=1 \mod{4}$
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Accepted

### Prove that if $3\mid a^2+b^2$ then $3\mid a$ and $3\mid b$.

The original question was to prove that $c\mid a^2+b^2$ implies $c\mid a$ and $c\mid b$, which as many answers show isn't true. But this is true if you take the assumption that there isn't a square ...
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### $xy$ is a quadratic residue mod $p$ iff $x$ is a quadratic residue mod $p$, where $y$ be a (nonzero) quadratic residue mod $p$

Suppose that $xy$ is a QR of $p$. Then $xy\equiv z^2\pmod{p}$ for some $z$. Since $y$ is a QR of $p$, we have $y\equiv w^2\pmod{p}$ for some $w$. Thus $xw^2\equiv z^2\pmod{p}$. Multiply both sides ...
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### Proof that square of an $\sqrt{2 - 4q}$ is not an integer

Apart from @Vincent's answer, here's another way of looking at it $$x=\sqrt{2-4q}$$ Clearly, $2-4q$ is divisible by $2$ but not by $4$ for any integer $q$. Thus, it cannot be a perfect square.
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### Is there any simple trick to solve the congruence $a^{24}\equiv6a+2\pmod{13}$?

"I am interested in Theorem statement, corollary, or Trick or Logic which solves this problem within one minute." Ok, so perhaps you are looking at Fermat's Little Theorem, where $n$ is prime, and $a$...
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• 17.8k
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### Do there exist Artificial Squares?

Suppose that $E = p^{2k+1} n$ where $(n,p) = 1$. Take $w = p^{2k+2}$. If $E$ is a quadratic residue modulo $p^{2k+2}$ then there exists $m$ such that $$p^{2k+2} \mid p^{2k+1} n - m^2.$$ In ...
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### Is there any simple trick to solve the congruence $a^{24}\equiv6a+2\pmod{13}$?

By Fermat's little theorem, $a^{24} = (a^{12})^2 \equiv 1^2 = 1 \pmod{13}$ (assuming $13\nmid a$, which is the case here).Thus you're really trying to solve $1\equiv6a+2\pmod{13}$, which is much ...
• 64.3k
Accepted

### Proof that square of an $\sqrt{2 - 4q}$ is not an integer

This answer builds completely on GoodDeed's excellent answer, but I think going into further detail is useful. Consider that $$x=\sqrt{2-4q} = \sqrt{2 (1 - 2q)} = \sqrt 2 \sqrt{1 - 2q}$$ $1 - 2 q$ ...
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### Prove that if $3\mid a^2+b^2$ then $3\mid a$ and $3\mid b$.

This is not true. Take, for example, $a=3$, $b=4$ and $c=5$.
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Accepted

### For a prime $p\ge 17$ is $\dfrac{p^2-1}{24}$ ever a prime?

When $p-1>24$, then $24$ can't cancel any of the factors $p-1$ and $p+1$ properly, so the number $\frac{p^2-1}{24}$ is always composite for $p-1>24$, other cases can be calculate manually...
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### If $3|(a^2 + b^2)$, show that $3|a$ and $3|b$.
The key point is that if $3$ does not divide $x$, then $x^2$ leaves remainder $1$, because $x$ is of the form $3k\pm1$. Therefore: If $3$ divides $a$ but not $b$ (or vice-versa), then $a^2+b^2$ ...