Conjecture: There exists no number k s.t. k^2+2 is a multiple of 7

1. Oct 20, 2012

middleCmusic

Hey everyone,

I was doing a problem in my Discrete Mathematics book and it called for finding an infinite number of counterexamples to the statement "7n+2" is a perfect square (which fails for n=3 at least).

In my search for such an infinite counterexample, I tried to find A, n=n(k,A), such that 7n+2=49k2-A2 for some A, and for all k. Then, I would be able to say that for n of the form, n=(k,A), where A is some fixed number, 7n+2=49k2-A2, which is a perfect square (7k)2 minus the fixed number A2. Clearly, if you can find this then you're done because squares will get large enough that the distance between them is more than A2, and thus, for a large enough n, 7n+2=49k2-A2 would not be a perfect square, as the next lowest square would be more than A2 away.

Anyway, in my search for this number, I let n be a quadratic function of k, n = xk2+yk+z for some fixed integers x, y, z to be determined, and from there I got these equations:

7(xk2+yk+z)+2=49k2-A2, which gives

7x=49 => x=7
yk=0 => y=0
7z+2=-A^2 => z=(-A2-2)/7

Now the first two equations give quick answers for x and y, but in order for 7 to be an integer, A2+2 must be a multiple of 7.

This is where I was led to the current title of this thread.

Now, I've put the first 1000 numbers of the form A2+2 (i.e. A=1, 2, 3, ..., 1000) into Excel and not a single one of them has been divisible by 7. Since, statistically, every 7th number should be divisible by 7, if we consider A2+2 to be a fairly random number, I thought that there must be something odd going on here that none of them had been a multiple of 7.

I could continue to put numbers into Excel (checking the first 2000, 5000, 10 000, and so on), but I thought I'd ask people here if they thought there was a simple number theory explanation for this. It's particularly pertinent for me since my whole proof hinges on an integer z existing. And if it doesn't, I shall have to try another route.

Thanks,

Chris

P. S. Also, if this statement is true (my conjecture), then it is either something that has already been proved or a pretty interesting result (to me at least), so I'd appreciate it if someone was able to prove the conjecture or point out where it's already been proven.

2. Oct 20, 2012

awkward

If $x \equiv y \pmod{7}$, then $x^2 +2 \equiv y^2+2 \pmod{7}$.

Therefore, if you check and find $x^2 + 2 \not \equiv 0 \pmod{7}$ for x = 0, 1, 2, .., 6, then you have shown that $x^2 +2 \not \equiv 0 \pmod{7}$ for all integers x.

3. Oct 20, 2012

middleCmusic

Thanks, awkward! You just saved me a lot of time making charts in Excel.

I had tried to go in the opposite direction, but obviously that doesn't work so well because the cancellation law with respect to modular arithmetic only works when the cancelled number is relatively prime to the mod. But your proof is awesome.

I was also worried because I know that conjectures are not welcomed here (in general), but I figured that something like this is not what the moderators had in mind when they banned "wild conjectures", right? I figure that an algebraic statement which is easily proven or disproven is much more timid than "I totally proved that Einstein was wrong about everything!"

4. Oct 20, 2012

Staff: Mentor

It asks for proof by induction, doesn't it?

5. Oct 20, 2012

Bacle2

If n is a number ending in 5, 7n ends in 5, and 7n+2 ends in 7. Same for n ending in

3--7n+2 also ends in 3.

6. Oct 20, 2012

ramsey2879

I don't see what k^2 + 2 has to do with the textbook problem so I don't understand your post. A better way to attack the textbook problem would be to consider what values n^2 can have mod 10 and find x mod 10 such that 7x + 2 mod 10 does not equal the possible values of n^2 mod 10. Do this and an infinite number of integers = x mod 10 are counter examples. See Bacle2's post.