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Almost Composite

If p and q are primes, and x2 – px + q = 0 has distinct positive integer roots, find p and q.

Because x^2-px+q=0 is quadratic, it must have exactly two distinct roots; let’s call these roots x_1, x_2.

Since x_1,x_2 are positive integer solutions to x^2-px+q=0, then it must be the case that

    \begin{align*} (x - x_1)(x - x_2) &= x^2-px+q \\ x^2-(x_1 + x_2)x + x_1x_2 &= x^2-px+q \\ &\iff p = x_1 + x_2, q = x_1x_2 \end{align*}

Now, since q is prime, then in order for q=x_1x_2, one of the two roots must be 1 or else q becomes composite. Suppose x_1 = 1, then q = x_2 is prime, and p = q + 1 is also prime. Hence, p,q are consecutive primes.

Due to the fact that all consecutive pairs must have exactly one even (even-odd or odd-even), then a consecutive pair of primes must contain a two. The only satisfying pair then is q = 2, p = 3, giving us the quadratic

    \[x^2-3x+2 = 0\]

with solutions 1 and 2. \square

Posted By Lee

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