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If $a, b$ are positive integers, does $\;b\mid(a^2 + 1)\implies b\mid (a^4 + 1)\;$?

Explain if this is true or not. If no, give a counterexample.

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There's no equal sign or inequality here. –  Ataraxia Jun 7 '13 at 0:31
    
It's not supposed to have equal signs, it's basically asking does it follow from "p" that "q" is true –  David Jun 7 '13 at 0:32
    
If possible can you explain when this may be true? –  David Jun 7 '13 at 0:34
    
Sorry, the "/" was throwing me off. –  Ataraxia Jun 7 '13 at 0:35
1  
What you have stated is not true. If you want it to be a true statement, try showing that $ b | (a^2 + 1) \Rightarrow b | (a^4 - 1)$. –  Calvin Lin Jun 7 '13 at 0:55

3 Answers 3

up vote 6 down vote accepted

Suppose $b = 5, a = 3$:

$$5\mid (3^2 + 1)\;\;\text{but}\;\;5 \not\mid (3^4 + 1)$$

Or, simply choose $a = 2, b= 5$ and again, $$5\mid (2^2 + 1)\;\;\text{but}\;\;5 \not\mid (2^4 + 1)$$


What is true is that $b\mid (a^2 + 1) \implies b\mid(a^2 + 1)^2$,

but note that $$\begin{align}\;(a^2 + 1)^2 &= (a^4 + 2a^2 + 1) \\ \\ &= \left[(a^4 + 1) + 2a^2\right]\\ \\ &\neq (a^4 + 1)\end{align}$$

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One example does the trick! :-) +1 –  Amzoti Jun 7 '13 at 0:33
    
Exactly what's needed! –  Sami Ben Romdhane Jan 18 at 14:33

Hint: $\,\gcd(a^2\!+\!1,a^4\!+\!1) = \gcd(a^2\!+\!1,(a^2\!+\!1)(a^2\!-\!1) + 2) = \gcd(a^2\!+\!1,2)$

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If positive integer $b$ divides $a^2+1$ and $a^4+1$ for some integer $a$

$b$ must divide $(a^4+1)-(a^2-1)(a^2+1)=2$

So, $b=1$ or $2$

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