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

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

    How to solve this if ^denotes power ?
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    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 x+1... Your writing is not clear ~
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