Question

Prove that there are an infinite number of values for "n" such that n^2 - 3 is divisible by a square number.

Answer 1

Examples: 27^2-3 is divisible by 11^2. 61^2-3 is divisble by 13^2. 94^2-3 is divisible by 11^2. (Those are the only values of n between 1 and 100, I think.)

Answer 2

Number theory is tough :(

Answer 3

I'd lean toward a proof by contradiction on this one.

Answer 4

So I assume that there is a maximum value of "n" for which this is true, but then what? The divisor is often 11^2 (and sometimes 13^2), should I only consider that subset? Those with work 11^2 are n=27,94,148,215,169,336,390,457 (that's all upto 500).

Answer 5

Not really just a maximum, just a finite number of n's. I only know basic number theory, though. Can't help you much, sorry. :(

Answer 6

We've only done about a month of it, and it really is a tough area! Thanks for trying anyway :)

Answer 7

If anybody is interested, I found a solution using 11^2 :) I worked out that 27^2=3 (mod 121) and 3^5=1 (mod 121), and so 27^10=1 (mod 121). This means that 27^2=27^12=27^22=...=3, and so 27^(10x+3)-3 is always divisible by 121.

Answer 8

Wow. That's nice! That's actually a sufficient proof! =))

Answer 9

Thanks :) It's probably not what the textbook people wanted me to do, but it works so I'm happy :)

Answer 10

(121 n + 27)^2 - 3 for any n is divisible by 11^2

Answer 11

\[(121 n+27)^2-3 = 14641 n^2+6534 n+726\\

\frac { 14641 n^2+6534 n+726}{11^2}=121 n^2+54 n+6

\]

Answer 12

Ooh, your proof is even nicer than mine :D Thanks for sharing!

Answer 13

Fact: Let k be a number such that k^2-3 is divisible by a^2, then
(a^2 n + k)^2 - 3 is divisible by a^2 for any n.
(a^2 n + k)^2 - 3 =a^4 n^2 + 2 a^2 n k +( k^2 -3)
Every term is divisible by a^2 so the sum is.