Confusion from prime

Number Theory Level 3

Find the sum of all prime number(s) of $p$ such that $17p+1$ is a perfect square.

The answer is 19.

1 solution

Brian Charlesworth
Aug 8, 2015

We require that $17p + 1 = n^{2}$ for some integer $n$ and prime $p.$ (Without loss of generality we can assume that $n$ is positive.) The equation can be rewritten as

$17p = n^{2} - 1 \Longrightarrow 17p = (n - 1)(n + 1).$

Now since both $17$ and $p$ are prime, by the Fundamental Theorem of Arithmetic we can either have

(i) $n - 1 = 1, n + 1 = 17p \Longrightarrow n = 2, 17p = 3,$ which is not possible,
(ii) $n - 1 = p, n + 1 = 17 \Longrightarrow n = 16,$ but then $p = 15$ is not prime, or
(iii) $n - 1 = 17, n + 1 = 19 \Longrightarrow n = 18$ and $p = 19,$ which is prime.

The only possible solution is thus $p = \boxed{19}.$

Nice solution

Nasim Bappi - 5 years, 10 months ago

You missed the case $n-1=p, n+1=17$ , which of course doesn't work, since $15$ isn't prime.

mathh mathh - 5 years, 10 months ago

Thanks; edit made. I shouldn't have dismissed that case without first mentioning it.

Brian Charlesworth - 5 years, 10 months ago

You can simply say that 17p is a product of 2 numbers whose difference is 2. So the numbers can't be 17p and 1 because then p will not be an integer. So the two numbers are 17 and p. So p is either 15 or 19. 15 is composite so answer is 19.

Kushagra Sahni - 5 years, 10 months ago

I want to know on what basis have you considered p a prime number as in the question it is not mentioned that p is a prime number??? Please do reply as I want to know!!

이채 린 - 5 years, 10 months ago

The question does actually ask for "all prime number(s) of $p$ ". If we were asked for positive integers in general, there would be an infinite number of solutions. In this more general case, we would be looking again for $n$ such that

$17n + 1 = m^{2} \Longrightarrow 17n = m^{2} - 1 = (m - 1)(m + 1)$

for some (positive) integer $m.$ By letting $m = 17k + 1$ for some positive integer $k,$ our equation would then become

$17n = (17k)(17k + 2) \Longrightarrow n = k(17k + 2)$ for some positive integer $k.$

This would give us an infinite number of possible values $n$ such that $17n + 1$ is a perfect square, starting with $19, 72, 159, .....$

We could also let $m = 17k + 16,$ yielding the equation

$17n = (m - 1)(m + 1) = (17k + 15)(17k + 17)$

$\Longrightarrow n = (17k + 15)(k + 1) = 17k^{2} + 32k + 15.$

This gives us another infinite "family" of solutions, starting with $64, 147, 264, ....$

Yet from both of these "families" of solutions, $19$ is still the only prime. :)

Brian Charlesworth - 5 years, 10 months ago

oh ok I thought of that for a second but needed an evaluation thank you for clearing my confusion. :)

이채 린 - 5 years, 10 months ago

Confusion from prime

The answer is 19.

1 solution

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