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Context: this question is from an old Australian Maths Olympiad and I was doing revision when I came across this question.

Prove that there exists no positive integers $a,b,c,d$ such that $$\begin{cases} b^2=10a-1\\ c^2=13a-1\\ d^2=85a-1 \end{cases}$$

By rearranging the first equation, we have $a=\frac{b^2+1}{10} \implies b^2 \equiv 9_{\bmod 10}$, if we assume that $b\neq 3$ where $(a,b)=(3,1)$ is a solution for the first equation. Then only squares that end in 9 are $b\equiv 3_{\bmod10}$ or $b\equiv 7_{\bmod10}$. By continuing we can find $a$ in terms of $k$ where $b=10k+3$ or $10k+7$.

John O'neil
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    In equation 1, $b = 3, a = 1$ is a solution, and as you'll note, $b$ is not congruent to 9 $\bmod 10$. Perhaps you'd like to edit that part of your claim, and fill in the details of the remaining sentence. – John Hughes Mar 22 '25 at 00:19
  • Where did this question come from. Context please [tour] – Mike Mar 22 '25 at 01:18
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    @Mike Thanks for pointing that out. I've added it in the question now. – John O'neil Mar 22 '25 at 01:20
  • Oh thanks @JohnOmielan I will fix this right now - the rest of the sentence was just for clarity. I'll consider removing it if it actually makes my effort more confusing. – John O'neil Mar 22 '25 at 04:28
  • @John I added an answer giving a simple, conceptual proof of a generalization. What method(s) are in the official solution? $\ \ $ – Bill Dubuque Mar 26 '25 at 17:09

3 Answers3

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I don't see how to continue from what you've done. Instead, adding the first equation multiplied by $-24$, the second one by $25$, and the third one by $-1$, gives

$$\begin{equation}\begin{aligned} -24b^2 + 25c^2 - d^2 & = -24(10)a + 24 + 25(13)a - 25 - 85a + 1 \\ -24b^2 + 25c^2 - d^2 & = (-240 + 325 - 85)a + (24 - 25 + 1) \\ -24b^2 + 25c^2 - d^2 & = 0 \\ 24b^2 + d^2 & = 25c^2 \end{aligned}\end{equation}$$

Note that, as Bill's comment indicates, an easier way to get the above result is by subtracting the middle original equation from the first and last ones, then use the first result in the second one, to arrive at

$$c^2 - b^2 = {\color{red}{3a}}, \;\; {\color{green}{d^2 - c^2}} = {\color{red}{24\cdot 3a}} = {\color{green}{24(c^2 - b^2)}}$$

From the first original equation, we get that $b$ is odd. Also, if $a$ were even, then $b^2 \equiv 3\pmod{4}$, which is not possible. Thus, $a$ must be odd, so $c^2 = 13a - 1$ and $d^2 = 85a - 1$ are both even, so both $c$ and $d$ are even. As Will's comment points out, we have $24b^2 \equiv 8 \pmod{16}$. However, $c^2$ and $d^2$ are congruent to $0$ or $4$ modulo $16$, so $25c^2 - d^2 \equiv 9c^2 - d^2 \pmod{16}$ becomes $9(0) - 0 \equiv 0 \pmod{16}$, $9(4) - 0 \equiv 4 \pmod{16}$, $9(0) - 4 \equiv 12 \pmod{16}$ or $9(4) - 4 \equiv 0 \pmod{16}$, i.e., $d^2 - 25c^2 \equiv 0, 4, 12 \pmod{16}$. This congruences list not including $8$ shows $24b^2 \neq 25c^2 - d^2$, i.e., no integer solutions exist.

Originally, I used that the resulting equation gives $d^2 \equiv c^2 \pmod{8} \;\to\; (d - c)(d + c) \equiv 0 \pmod{8}$, so $c$ and $d$ differ by a multiple of $4$ (since if $c$ and $d$ differed by an odd multiple of $2$, then both $d - c$ and $d + c$ would have only one factor of $2$). Also, since the $3$ original equations show that $b \lt c \lt d$, there exists an odd positive integer $e$, and a positive integer $f$, such that

$$b = c - e, \;\; d = c + 4f$$

Substituting these into our earlier equation results in

$$\begin{equation}\begin{aligned} 24(c - e)^2 + (c + 4f)^2 & = 25c^2 \\ 24c^2 - 48ce + 24e^2 + c^2 + 8cf + 16f^2 & = 25c^2 \\ - 48ce + 8cf & = - 24e^2 - 16f^2 \\ c(-6e + f) & = -3e^2 - 2f^2 \end{aligned}\end{equation}$$

However, as $c$ is even, the left side is even, but $e$ being odd means the right side is odd! This contradiction shows there are no integer solutions to the original set of equations.

John Omielan
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    I'll look more in the morning; once you reach $24b^2 + d^2 = 25 c^2$ with $b$ odd and $c,d$ even, then $24 b^2 \equiv 8 \pmod {16}$ while $d^2 - 25 c^2 \equiv 0, 4, 12 \pmod {16}$ – Will Jagy Mar 22 '25 at 04:29
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    @WillJagy Thank you for the feedback. I've added your suggestion to my answer (although, technically, it's $25c^2 - d^2$ to check on). I thought there should be some direct modular argument, but I made a mistake originally when I checked earlier, so I didn't see it then. Nonetheless, in hindsight, the approach I used does basically indicate that an appropriate power of $2$ modulo check should work. – John Omielan Mar 22 '25 at 04:50
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    Clearer & easier: $ $ eliminate $,\color{#c00}a,$ by add/subtract middle to first and last OP equations: $$\begin{align} c^2 - b^2 &= \color{#c00}{3a}\[.3em] \color{#0a0}{d^2 - c^2} &= \color{#c00}{3a}\cdot 24 = \color{#0a0}{(c^2-b^2),24}\end{align}\qquad\qquad$$ You are welcome to use this in the answer. $\ \ $ – Bill Dubuque Mar 22 '25 at 17:22
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Lemma $ $ A $\rm\color{#0a0}{difference}$ of $\rm\color{#0af}{even}$ squares is $\not\equiv \color{#c00}8\pmod{\!16},\,$ else cancelling $\:\!4\:\!$ yields a difference of squares $\equiv \color{f40}2\pmod{\!4},\,$ contra squares $\equiv\:\! 0,1\pmod{\!4},\, $ cf. unachievable mod sums.

Theorem $ $ The system of equations below has no solutions in integers $\,a,b,c,d\,$ when $\,j,k_i,\ell_i,n\,$ are integers with $\:\!k_i,\:\!\ell_i$ odd; $\ n\equiv_4 \color{darkorange}3;\,$ $\ k_2\!-\!k_1\equiv_{16} \color{#c00}8,\,\ \ell_1\!\equiv_{16} \ell_2$ $$\begin{align} [\![1]\!]&\,\quad b^2=\:\! 2j\ a+n\:\!\ \ \ \,[= 10a-1\ \ \rm in\ OP\:\!]\\ [\![2]\!]&\,\quad c^2=\:\! k_1\ a+\ell_1\, \ \ [= 13a-1\ \ \rm in\ OP\:\!]\\ [\![3]\!]&\,\quad d^2\!=\:\! k_2\ a+\ell_2\, \ \ [= 85a-1\ \ \rm in\ OP\:\!] \end{align}\qquad\qquad\ $$

Proof $\ a\,$ is odd (else $\,[\![1]\!]\Rightarrow b^2\equiv_4 \color{darkorange}3\:\!)\:$ so $\:a,k_i,\ell_i\:\!$ odd $\,\Rightarrow c,d\,$ $\rm\color{#0af}{even}$ by $[\![2]\!],[\![3]\!]$

so $\,[\![3]\!]-[\![2]\!]\, \Rightarrow\, \bmod 16\!:\ \color{#0a0}{d^2-c^2} \equiv \color{#c00}8a\equiv \color{#c00}8,\,$ by $\,a\,$ odd, contra Lemma. $\ \bf\small QED$

Bill Dubuque
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  • We could replace $[![1]!]$ by "$a$ is odd" since that is the only role it plays in the proof. Then it becomes much clearer that it boils down to the Lemma. $\ \ $ – Bill Dubuque Mar 26 '25 at 19:28
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Modulo $16$ one has $$\begin{cases} b^2=10a-1\\ c^2=-3a-1\\ d^2=5a-1 \end{cases}$$ Since $b$ should be odd and $\boxed{(\mathbb Z/16\mathbb Z)^2=\{1,4,9,0\}}$, we have $$b^2=10a-1=\begin{cases}1\\9\end{cases}\implies\begin{cases}10a-2=0\\10a-10=0\end{cases}$$ It follows $2(5a-1)\equiv0\pmod{16}$ and $10(a-1)\equiv0\pmod{16}$ so we have two values modulo $16$: $a=5$ and $a=1$.

With $a=5$ we would have $d^2=25-1=24\equiv8$ and with $a=1$ we would have $c^2=-3-1=-4=12$ and both cases are not squares in $\mathbb Z/16\mathbb Z$. We are done.

Ataulfo
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  • Very nice. I'm slightly surprised that we didn't need to use $d^2$. – Calvin Lin Mar 22 '25 at 09:28
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    Wait... we have $a\equiv 1,5 \pmod{8}$ so $a \equiv 1,5,9,13 \pmod{16}$ but if $a \equiv 5,9 \pmod{16}$ then $c^2 \equiv 0,4 \pmod{16}$, so we aren't done, are we? – Evariste Mar 22 '25 at 09:48
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    @CalvinLin We need it to conclude $a \equiv 1,13 \pmod{16}$ by my comment above (then the reasoning would work I guess) – Evariste Mar 22 '25 at 10:45
  • @Evariste: All the reasoning is modulo $16$. In $\mathbb Z/16\mathbb Z$ you can use the sign $=$ instead of $\equiv$. That's what I did using $=$. – Ataulfo Mar 22 '25 at 14:08
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    As mentioned, you argument has gaps - cases $,a\equiv 5.9\pmod{16}.,$ As such, it is best to delete the answer until you can remedy that. $\ \ $ – Bill Dubuque Mar 22 '25 at 14:54
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    $2(5a-1)\equiv 0 \pmod {16}$ and $10(a-1)\equiv 0 \pmod {16}$ does not imply $a\equiv 1, 13 \pmod {16}$. The first equation have also the solution $a\equiv 5 \pmod{16}$ and the second equation have also the solution $a\equiv 9 \pmod{16}$. – jjagmath Mar 22 '25 at 18:00
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    Your edit doesn't close the gap since it still excludes only two of the above $4$ listed solutions $,a\equiv 1,5,9,13\pmod{!16}.\ \ $ – Bill Dubuque Mar 23 '25 at 02:36