# looking for integer pairs $(a,c)$ such that $4a^2 + 8c^2 - 4c + 1$ is a perfect square

Hi. I'm looking to find integer pairs $(a,c)$ such that $4a^2 + 8c^2 - 4c +1$ is a perfect square.

The sum is odd so I set the sum equal to $(2n+1)^2$ to cancel out the 1s and I end up with $\boxed{a^2 + 2c^2 - c = n^2 + n}$.

I haven't found a valid technique to break this down yet but I think it could be a pell's equation in disguise. I'm wondering if anyone can give me some insight on this problem. Thanks.

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Is this homework in an elementary number theory course? Try math.stackexchange.com – Robert Israel Oct 10 '11 at 19:37
Given GH's answer, this does not look too elementary to me. Are we missing something? – Igor Rivin Oct 10 '11 at 19:39
similar to mathoverflow.net/questions/77207 and the answers give the two main constructions, just as for Pythagorean triples, one way (Pietro) is a parametrization, another way (GH) is to find the full orthogonal group of an indefinite ternary form. As the OP is a high-school student, Pietro's method is the way to go. – Will Jagy Oct 10 '11 at 20:20
Thanks everybody for your kind comments. I really appreciate it. – Junichi Oct 10 '11 at 20:29

## 3 Answers

Am I missing something? If you write

$$8c^2 - 4c + 1 = n^2 - 4a^2 = (n-2a)(n+2a)$$

then you can let $c$ be anything, let $hk$ be any factorization of $8c^2-4c+1$, and get $a=(k-h)/4$ and $n=(k+h)/2$. (Note that $h$ and $k$ are necessarily congruent mod 4, so $a$ is an integer.)

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@Pietro, sorry, I was taking $n$ to be the perfect square equal to $4a^2 + 8c^2 - 4c +1$, not the $n$ from the OP's $(2n+1)^2$. Maybe I should have used a capital $N$ to avoid confusion. Note that since $h$ and $k$ are both odd and congruent mod 4, you get (my) $n$, $(k+h)/2$, to be odd. – Barry Cipra Oct 10 '11 at 22:48
Is your $n$ that one? For the equation $a^2+2c^2−c=n^2+n$ I got $a=(k−h)/4$ and $n=(k+h+2)/4$, with the same meaning of $h$ and $k$. – Pietro Majer Oct 10 '11 at 22:49
Oops, I should have just said I was taking $n^2$ (not $n$) to be the perfect square equal to $4a^2 + etc.$ – Barry Cipra Oct 10 '11 at 22:50
Ok I see ! – Pietro Majer Oct 10 '11 at 22:50
@Pietro, amusingly when I reviewed the comments here just now, mathoverflow says your original comment/question was posted "8 mins ago" and my reply to it was posted "9 mins ago"! – Barry Cipra Oct 10 '11 at 23:03

With the substitution $x:=2a$, $y:=4c-1$, $z:=2n+1$ your framed equation becomes $$2x^2+y^2-2z^2=-1.$$ Equations of this type are studied thoroughly in Section 13.6 "Representation by Anisotropic Ternaries" of Cassels: Rational Quadratic Forms (Academic Press, 1978). In particular, Theorem 6.2 on p.305 is directly applicable here: every integral solution is the image of a reduced solution by an automorph of $2x^2+y^2-2z^2$, and reduced integral representations can be found effectively.

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Thanks for the quick response! Does that imply that we can't generate solutions to this with elementary methods(I'm a high school student which is why I'm wondering this)? Also I'm wondering if I can find these integral representations online. Thanks! – Junichi Oct 10 '11 at 19:49
No. This is a problem that I encountered while doing my research project. My original problem is to find triplets (a,b,c) such that $a+(a+1)+(a+2)+...+(a+b)=a+b+1+(a+b+2)+...+(a+b+c)$. – Junichi Oct 10 '11 at 19:55
There are infinitely many representations and they can be described explicitly and in elementary terms (while the proof certainly is beyond the high school level). But don't ask me to give the representations myself, since for that I would need to follow Cassels' approach thoroughly and do numerical work. There are experts here on quadratic forms, perhaps they can do this for you, I just wanted to provide a pointer. In general ternary quadratic forms are tricky, but indefinite ones are easier than definite ones. – GH from MO Oct 10 '11 at 19:58
Thanks a lot! Should I start another "question" regarding the equation type you gave me or should I simply look for information regarding this field myself? Again thank you so much for your help I've been stuck on this for days I really really appreciate it – Junichi Oct 10 '11 at 20:02
If you want to solve your problem you need to study indefinite ternary quadratic forms, e.g. from Cassels' book. You can use your problem to motivate yourself to learn lots of beautiful mathematics. Serre's classic, "A course in arithmetic", also contains an excellent introduction to this field. In general, I think, for high school students and also for undergraduate university students it is more important to learn mathematics than (to try) to do original research. – GH from MO Oct 10 '11 at 20:10

How about trying $4a^2+8c^2-4c+1=(2a)^2+(2c-1)^2+(2c)^2=n^2$ for some integer $n$? A simple one would be $12^2+3^2+4^2=13^2$. Hence the pair $(a, c)=(6, 2)$ works.

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