Prove that the equation $a^2+b^2=c^2+3$ has infinitely many integer solutions $(a,b,c)$.

Question

Prove that the equation $a^2+b^2=c^2+3$ has infinitely many integer solutions $(a,b,c)$.

My attempt:

$(a+1)(a-1)+(b+1)(b-1)=c^2+1$
This form didn't help so I thought of $\mod 3$, but that didn't help either. Please help. Thank you.

Answer 1

If $(a,b,c)$ is a solution to $a^2 + b^2 - c^2 = K,$ where your case is $K=3,$ then you get another solution with $$ (a+2b+2c, 2a+b+2c, 2a+2b+3c). $$ Starting with $a,b,c>0,$ you can apply this again and again, the entries grow, forever.

           a           b           c
           2           0           1
           4           6           7
          30          28          41
         168         170         239
         986         984        1393
        5740        5742        8119
       33462       33460       47321
      195024      195026      275807
     1136690     1136688     1607521
     6625108     6625110     9369319
    38613966    38613964    54608393
jagy@phobeusjunior

See http://en.wikipedia.org/wiki/Tree_of_primitive_Pythagorean_triples

The matrix calculation that shows that the value of my letter $K$ does not matter is $$ \left( \begin{array}{rrr} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 3 \end{array} \right) \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{array} \right) \left( \begin{array}{rrr} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 3 \end{array} \right) = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{array} \right) $$ In the matrices with the 2's and 3's, usually the one on the left is the transpose of the one on the right in the multiplication shown, but this time they are symmetric so the transpose is the same. Anyway, for any $K,$ you just need to find one solution $(a,b,c)$ with $a,b,c \geq 0$ and, for preference, $\gcd(a,b,c) = 1,$ and then this process gives infinitely many positive solutions for the same $K,$ and they stay primitive as well (gcd 1).

This can be written as a single equation which is fairly readable in this case, $$ \color{blue}{ (a+2b+2c)^2 + (2a+b+2c)^2 - ( 2a+2b+3c)^2 \; = \; a^2 + b^2 - c^2}. $$

Answer 2

In order that an integer $n$ can be represented as $c^2-b^2$, it suffices that $n\not\equiv{2}\pmod{4}$. In such a case $n$ can be written as the product of two divisors with the same parity, $n=pq$, and we can take $c=\frac{p+q}{2},d=\frac{p-q}{2}$. So it is sufficient to prove that for an infinite number of integers $a$, $a^2-3\not\equiv 2\pmod{4}$. This happens every time $a$ is even.

For instance, take $a=10$. We can write $n=a^2-3=97$ as $1\cdot 97$, hence $$ (a,b,c) = (10,48,49) $$ is a solution of $a^2-3=c^2-b^2$.

Answer 3

Put $a=6k^2-2$, $b=6k$ and $c=6k^2+1$. Then

$$\left(6k^2-2\right)^2+(6k)^2=\left(6k^2+1\right)^2+3.$$

Behind these solutions is the observation that $$\left(\frac{x+y}{2}\right)^2-\left(\frac{x-y}{2}\right)^2=xy$$ We can re-write the given equation as $c^2-a^2=b^2-3$ and choose $b$ such that $b^2-3$ is the product of two odd numbers $xy$. This allows us to choose $a=\frac{x-y}{2}$ and $c=\frac{x+y}{2}$.

Answer 4

Although the formula I posted. Integral solutions of $x^2+y^2+1=z^2$

Thought and decided to write as simple as possible. In equation:

$$a^2+b^2=c^2+q$$

$q$ - is specified by the problem statement. Then choose one solution $a$ so that it was possible to factor this way.

$$a^2-q=t(t+2s)$$

So the solution of this equation is always there. And can be written as:

$$b=s$$

$$c=s+t$$

Answer 5

Note that the requirement is simply to demonstrate infinitely many solutions - we are not required to find all solutions.

Setting $c=b+1$, we see that $a^2 = c^2-b^2+3 = 2b+4$

Therefore for any even $a>2$, we can choose $b=\frac{a^2-4}{2}$ and $c=b+1$. This gives infinitely many solutions as required.