Let $a^2,b^2,c^2$ be natural numbers in an arithmetic progression with difference $k$.show that $24\mid k$

Question

I need to solve the following question: Let $a^2,b^2,c^2$ be natural numbers in an arithmetic progression, with difference $k$. Show that $24\mid k$.

I'm just learning the basics of number theory right now, and I struggle with how to even start the question. Would greatly appreciate guidance on how to try and solve this problem, and similar ones.

Thanks.

Answer 1

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

$(2b)^2=(a+c)^2+(a-c)^2$

Using https://en.m.wikipedia.org/wiki/Pythagorean_triple,

$a+c=p(m^2-n^2)$

$a-c=2pmn$

$2b=p(m^2+n^2)$

where $m,n$ are co-primes with opposite parity

$\implies p$ is even

So, it's sufficient to show $48|(a^2-c^2)$

$\iff48|2p^2mn(m^2-n^2)$

As $p$ is even, this $\iff6|mn(m^2-n^2)$

As $m,n$ have opposite parity, one of them is even

If $3\nmid mn$ can you prove $3|(m^2-n^2)$

Answer 2

On how to start

You should first begin asking yourself how to use the information you're given. You might need to take advantage of every detail - there's no extra information. Never - or at least almost never.

A possible beginning would be, for instance, to notice that

\begin{align*} a^2+k&=b^2\\ \implies k&=b^2-a^2\\ &=(b+a)(b-a)\\ a^2+2k&=c^2\\ \implies k&=c^2-b^2\\ &=(c+b)(c-b) \end{align*}

You can now work with the expression in order to prove separately that $8\mid k$ and $3\mid k$.

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Observe for instance that since $$\frac{a^2+c^2}{2}=b^2\iff a^2+c^2=2b^2$$ $a,b,c$ have to have the same parity. Hence $$k=\underbrace{(b+a)}_{\text{even}}\cdot \underbrace{(b-a)}_{\text{even}}\implies 4\mid k$$

Answer 3

Although the previous answer is clever, given this problem I would attack it by considering things mod $3$ and mod $8$. In either mod, the sequence will still be in arithmetic progression (except that it is permitted to cross the $n-1 \rightarrow 0$ step; for example, in mod $8$, the sequence $3,6,1$ is in arithmetic progression).

In mod $3$, the only quadratic residues (numbers mod $3$ that are squares of other numbers mod $3$) are $0$ and $1$. So the sequence has to be $0,0,0$ or $1,1,1$; sequences like $0,1,2$ and $0,2,1$ contain the forbidden $2$. In either of those cases, $k \equiv 0 \pmod 3$.

In mod $8$, the only quadratic residues are $0, 1, 4$. So the sequence $a,b,c$ has to be one of: $(0,0,0), (1,1,1), (4,4,4), (0,4,0), (4,0,4)$.

The first three of those have $k \equiv 0 \pmod 8$. THe other two have $k \equiv 4 \pmod 8$ and need to be eliminated for our result to be proven.

The fourth sequence turns out not to work: It would mean $a$ and $c$ are both divisible by $4$ and $b$ is of the form $4n+2$ in order for $b^2\equiv 2 \pmod 8$. Then if $a=4m$ and $c=4p$, $$ (4m)^2 = (fn+2)^2 - 4 = 16n^2+16mn \\ b^2 = (4n+2)^2 + 4 = (16n^2 + 16mn+4) + 4 = (4p)^2 \\ 16n^2 + 16mn+8 = 16p^2 \\ 2(m^2+n^2) + 1 = 2p^2 $$ But the LHS is odd and the RHS is even, so the residue sequence $(0,4,0)$ does not work.

A similar argument shows $(4,0,4)$ also fails in the same way. Thus one of the $k\equiv 0$ sequences must hold, therefore $k$ is divisible by $8$.

From which $k$ is divisible by $3\times 8 = 24$.

Answer 4

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

If $(a,c)=d,$ let $a/A=b/B=c/C=d$

$A^2+C^2=2B^2$

$A,C$ must have the same parity, so both must be odd as they are relatively prime

For odd $p,p^2\equiv1\pmod8$

$A^2+C^2\equiv2\pmod8$

$B^2\equiv1\pmod4\implies B$ is odd

As $(A,C)=1, 3$ can divide at most one of them

If $3$ divides $A,$

$A^2+C^2\equiv1\equiv-2\pmod3$

$\implies B^2\equiv-1\pmod3$

But for any integer $q,q^2\equiv0,1\pmod3$

So,$3\nmid A\implies3\nmid B$

$\implies (6,A)=(6,B)=(6,C)=1$

If $(6,r)=1,r=6t\pm1$

$r^2=24t^2+24\cdot\dfrac{t(t\pm1)}2+1\equiv1\pmod{24}$