Daniell Phillips

2022-01-07

Is there a pair of integers $\left(a,b\right)$ such that $a,{x}_{1},{y}_{1},b$ is part of an arithmetic sequences and $a,{x}_{2},{y}_{2},b$ is part of a geometric sequence with ${x}_{1},{x}_{2},{y}_{1},{y}_{2}$ all integers?

Jim Hunt

It is given that ${x}_{1},{x}_{2},{y}_{1},{y}_{2}$ are all integers.
$a,{x}_{1},{y}_{1},b$ is part of an arithmetic sequences.
Thus, ${x}_{1}-a={y}_{1}-{x}_{1}$ the common difference is same in arithmetic sequences.
${x}_{1}-a={y}_{1}-{x}_{1}$
$2{x}_{1}={y}_{1}+a$
$a=2{x}_{1}-{y}_{1}$
So, $a$ is integer as ${x}_{1}$ and ${y}_{1}$ are integers. (1)
${y}_{1}-{x}_{1}=b-{y}_{1}$ the common difference is same in arithmetic sequences.
${y}_{1}-{x}_{1}=b-{y}_{1}$
$2{y}_{1}={x}_{1}+b$
$b=2{y}_{1}-{x}_{1}$
So, $b$ is integer as ${x}_{1}$ and ${y}_{1}$ are integers. (2)
It is given that $a,{x}_{2},{y}_{2},b$ is part of a geometric sequence. Hence, $\frac{{x}_{2}}{a}=\frac{{y}_{2}}{{x}_{2}}$ the common ratio is same in geometric sequence.
${x}_{2}{x}_{2}=a{y}_{2}$
$a=\frac{{x}_{2}^{2}}{{y}_{2}}$
${x}_{2}=\sqrt{a{y}_{2}}$
Similarly, ${y}_{2}=\sqrt{b{x}_{2}}$
Also, $\frac{{x}_{2}}{a}=\frac{b}{{y}_{2}}$ the common ratio is same in geometric sequence.
$ab={x}_{2}{y}_{2}$
$ab$ is an integer as ${x}_{2}$ and ${y}_{2}$ are integerand the product of two integer is an integer.
Thus, from equations it is proved that there is a pair of integers $\left(a,b\right)$.

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