Showing that one of three expressions is a perfect square

Question

$a$, $b$, $c$ are natural numbers such that

$a^2 + b^2 + c^2 = 1 + 2abc$

Prove that one of $\frac{a+1}{2}, \frac{b+1}{2}, \frac{c+1}{2} $ is a perfect square.

Since at least one of $a,b,c$ has to be odd, WLOG that $a = 2x-1$, where $x$ is a natural number. Then I tried to split the problem into the cases depending on the parity of $b,c$ but I haven't found anything yet.

Answer 1

If we solve the quadratic in $c$, we obtain

$$ c=ab \pm \sqrt{(a^2-1)(b^2-1)} \tag{1} $$

We see that $(a^2-1)(b^2-1)$ is a perfect square. Let $d \gt 0$ be the square-free kernel of $a^2-1$, so that $a^2-1=dA^2$ for some positive integer $A$. Then $b^2-1=dB^2$ for some positive integer $B$.

If $x_0^2-dy_0^2=1$ is the fundamental solution of $x^2-dy^2=1$, then we have two exponents $n$ and $m$ such that

$$ a+A\sqrt{d}=z_0^n, b+B\sqrt{d}=z_0^m \tag{2} $$ where $z_0=x_0+y_0\sqrt{d}$.

Multiplying or dividing the two relations above, we obtain

$$ ab-dAB +(aB+bA)\sqrt{d} = z_0^{n+m}, ab+dAB +(-aB+bA)\sqrt{d} = z_0^{n-m} \tag{3} $$

If $n$ is even, say $n=2p$ for some integer $p$, we can write $a+A\sqrt{d}=z_1^2$ where $z_1=z_0^p$. If we write $z_1=x_1+y_1\sqrt{d}$, then $x_1^2-dy_1^2=1$, and $a=x_1^2+dy_1^2=2x_1^2-1$, so $\frac{a+1}{2}=x_1^2$ is a perfect square.

Similarly, if $m$ is even we deduce that $\frac{b+1}{2}$ is a perfect square. Finally, if $n$ and $m$ are both odd, then $n\pm m$ are both even and we deduce from (3) and (1) that $\frac{c+1}{2}$ is a perfect square. This finishes the proof.

Answer 2

let me make this Community Wiki, people sometimes dislike computed lists; crucial, however, in exploring diophantine equations, looking for patterns.

Alright, there are always examples with repeats; for any positive $b,$ the triple with $a=1, c=b$ is a solution. Furthermore, Vieta JUmping takes this to $(2b^2 - 1, b,b),$

Here is a list of triples with small maximum, deliberately leaving out the solutions where one of the variables is equal to $1$ In all the triples I write in decreasing order

$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$

7   2   2
17   3   3
26   7   2
31   4   4
49   5   5
71   6   6
97   7   7
97   26   2
99   17   3
127   8   8
161   9   9
199   10   10
241   11   11
244   31   4
287   12   12
337   13   13
362   26   7
362   97   2
391   14   14
449   15   15
485   49   5
511   16   16
577   17   17
577   99   3
647   18   18
721   19   19
799   20   2
846   71   6
881   21   21
967   22   22
1057   23   23
1151   24   24
1249   25   25
1351   26   26
1351   97   7
1351   362   2
1457   27   27
1567   28   28
1681   29   29
1799   30   30
1921   31   31
1921   244   4
2024   127   8
2047   32   32
2177   33   33
2311   34   34
2449   35   35
2591   36   36
2737   37   37
2887   38   38
2889   161   9
3041   39   39
3199   40   40
3361   41   41
3363   99   17
3363   577   3

$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$