Value of $b$ in smallest triplet of perfect squares $(a,b,c)$ where $a+5k=b, b+5k=c.$

Question

Find the value of $b$ in the smallest triplet of perfect squares $(a,b,c)$ such that $a+5k=b$ and $b+5k=c$ for positive integers $a,b,c,k.$

I started by defining $a=a_1^2, b=b_1^2, c=c_1^2$ for convenience as they are perfect squares. Then, we know that $$a_1^2=a_1^2, \;(a_1^2+5k)=b_1^2, \;(a_1^2+10k)=c_1^2.$$ I'm not completely sure on how to continue from here, but I assume we have to solve the equaions in the problem and then try to minimize $b$ from there. May I have some help? Thanks in advance.

When I say smallest triplet, I mean a triplet that has its value of $b$ as minimal as possible.

Answer 1

You would like to have $p^2,q^2,r^2$ in arithmetic progression. So $$p^2+r^2=2q^2$$

Perhaps it is easier to catalog solutions to this Pythagorean-like equation. And then examine them for when there is a common difference that is a positive multiple of $5$.

Cataloging the solutions for these triples can be done in the same way as for Pythagorean triples. I leave out the details since that is a significant detour here. But solutions are

$$p=(u-v)^2-2v^2$$ $$q=u^2+v^2$$ $$r=(u-v)^2-2u^2$$

$\{u,v\}=\{0,j\}\implies(p^2,q^2,r^2)=(j^4,j^4,j^4)$ and differences are $0$.

$\{u,v\}=\{\pm j,\pm j\}\implies(p^2,q^2,r^2)=(4j^4,4j^4,4j^4)$ and differences are $0$.

$\{u,v\}=\{\pm1,\pm2\}\implies(p^2,q^2,r^2)=(1,25,49)$ and differences are $24$.

$\{u,v\}=\{\pm1,\pm3\}\implies(p^2,q^2,r^2)=(4,100,196)$ and differences are $96$.

$\{u,v\}=\{\pm2,\pm3\}\implies(p^2,q^2,r^2)=(49,169,289)$ and differences are $120$, finally a positive multiple of $5$.

And we know this is minimal for $b$ since we are listing these things in increasing order of $\sqrt{b}=u^2+v^2$.

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A few more:

$\{u,v\}=\{\pm1,\pm4\}\implies(p^2,q^2,r^2)=(49,289,529)$ and differences are $240$, also a positive multiple of $5$. (So a new solution not mentioned yet in comments or answers.)

$\{u,v\}=\{\pm2,\pm4\}\implies(p^2,q^2,r^2)=(16,400,784)$ and differences are $384$.

$\{u,v\}=\{\pm3,\pm4\}\implies(p^2,q^2,r^2)=(289,625,961)$ and differences are $336$.

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This is only cataloguing primitive solutions to that diophantine equation. A full catalog should also explore the non-primitve solutions. Like the solution above $(1,25,49)$, scaled by $25$ to get the original OP observed solution.

Answer 2

It depends on what you mean by minimal. The triple $25, 625,1225$ has a smaller $a$, but $49,169,289$ beats it for $b$ and $c$.

Answer 3

\begin{align*} a+5k&=b \tag{1}\\ b+5k=c \implies c-5k&=b \tag{2}\\ \implies a+c&=2b \tag{3}\\ \implies a=b=c&=1^2 \tag{4} \end{align*} This works only if $\space k = \dfrac{b - a}{5}\space$ and $\sqrt{2b-a}= \big\lfloor\sqrt{2b-a}\space\big\rfloor$ The first solution $(3)\rightarrow (4)$ is easy to see if we let $k=\dfrac{1-1}{5}=0\space $ but it is required that $k>0$ so the smallest triple for $a$ is $(1, 841, 1681) $ where $k=\dfrac{841-1}{5}=168$

The smallest triple for $b$ and $c$ and $k$ is $(49, 169, 289 )$ where $k=\dfrac{169-49}{5}=24$

These and other triples may be explored with:

 110 print “Enter HI number”; : input h1
 120 print "a","  b",,"  c",,"  k"
 130 for a1 = 1 to h1
 140    for b1 = 1 to h1
 150       k1 = (b1^2-a1^2)/5
 160       if k1 > 0
 170       if k1 = int(k1)
 180         c1 = 2*b1^2-a1^2
 190         if sqr(c1) = int(sqr(c1))
 200           print a1^2, b1^2, c1, k1
 210          endif
 220        endif
 230        endif
 240     next b1
 250  next a1