Find the least positive integer x such that |13|(x^2+1)

**gigglie** · May 5th 2008, 11:26 AM

How to solve this if ^denotes power ?

**Moo** · May 5th 2008, 11:35 AM

Hello,

I haven't found another method yet.

Do it by using the possible congruences of x to 13.

$x \equiv 0 [13] \Longrightarrow x^2+1 \equiv 1 [13] \not \equiv 0 [13]$

$x \equiv 1 [13] \Longrightarrow x^2+1 \equiv 2 [13] \not \equiv 0 [13]$

$x \equiv 2 [13] \Longrightarrow x^2+1 \equiv 5 [13] \not \equiv 0 [13]$

$x \equiv 3 [13] \Longrightarrow x^2+1 \equiv 10 [13] \not \equiv 0 [13]$

$x \equiv 4 [13] \Longrightarrow x^2+1 \equiv 4 [13] \not \equiv 0 [13]$

$x \equiv 5 [13] \Longrightarrow x^2+1 \equiv 0 [13]$

So it's 5

That is, assuming that you meant 13 divides x²+1... Your writing is not clear ~

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