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

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

How to solve this if ^denotes power ?

2. 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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# find the least positive integer x such that 13/(xÂ² 1)

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