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$a$, $b$, $c$ 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.

justadumbguy
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2 Answers2

5

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.

Ewan Delanoy
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  • As frequently someone wins me by time. I have write " It is not too hard show that are infinitely many solutions" in my comment above precisely for your viewpoint. I have put an upvote (worth the clarification). – Piquito Oct 31 '21 at 18:14
  • Could the downvoter please explain his or her reasons ? @Piquito thanks for your upvote. – Ewan Delanoy Oct 31 '21 at 18:17
  • It is quite an undesirable reality in MSE that there are people who downvote without explaining the reason. And the worst thing is that there are indications that sometimes they put it out of pure bad conscience (there is also the bad habit of putting upvotes to another answer just to lower someone who does not want their reputation to increase. This attitude seems stupid to me). – Piquito Oct 31 '21 at 18:28
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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

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

Will Jagy
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