Question

Find the general expression for three positive integers in Arithmetic Progression, and such that the sum of every two numbers is a perfect square.

Answer 1

Hint:We first assume that the three numbers are- \[\text{ }a-d,a,a+d\] . These three numbers are in arithmetic progression. We then consider all possible sums of two numbers from the three assumed numbers and equate each of them to a different perfect square, e.g. ${{p}^{2}}$ , ${{q}^{2}}$ etc. Thereafter we carry on further evaluations and derive a general expression for the three numbers.

Complete step-by-step answer:
In the given question, first let us assume that the three numbers are- \[\text{ }a-d,a,a+d\] , where, the first term is \[\text{ }a-d\] , the second term is \[a\] and the third term is \[a+d\] and \[d\] is the common difference of the A.P. Now according to the problem statement, the sum of every two numbers include the sum of the first and second i.e. \[2a-d\] , the sum of the first and third i.e. \[2a\] , the sum of second and third i.e. \[2a+d\] . Now let us assume \[2a-d={{p}^{2}}\] , \[2a={{q}^{2}}\] and \[2a+d={{r}^{2}}\] . Now, since \[2a-d\] , \[2a\] and \[2a+d\] are also in A.P, we can write
 \[{{r}^{2}}-{{q}^{2}}={{q}^{2}}-{{p}^{2}}\]
From the general equation of the A.P as mentioned above, let us modify it as,
 \[\left( r-q \right)\left( r+q \right)=\left( q-p \right)\left( q+p \right)\]
Dividing both sides, we get,
 \[\Rightarrow \dfrac{r-q}{q-p}=\dfrac{q+p}{r+q}\]
Now, since this ratio is constant, we can write this to be,
 \[\dfrac{r-q}{q-p}=\dfrac{q+p}{r+q}=\dfrac{x}{y}\] Where x and y are mere constants, resulting a constant ratio.
We can write \[\dfrac{r-q}{q-p}=\dfrac{x}{y}\] . Cross-multiplying, we get \[yr-yq=xq-xp\] . Rearranging the terms we get \[xp=\left( x+y \right)q-yr\] .
 \[\therefore p=\dfrac{\left( x+y \right)q-yr}{x}\text{ }\cdot \cdot \cdot \cdot \cdot \cdot \cdot \text{ eqn 1}\]
We can write \[\dfrac{q+p}{r+q}=\dfrac{x}{y}\] . Cross-multiplying, we get \[yq+yp=xr+xq\] . Rearranging the terms we get \[yp=\left( x-y \right)q+xr\] .
 \[\therefore p=\dfrac{\left( x-y \right)q+xr}{y}\text{ }\cdot \cdot \cdot \cdot \cdot \cdot \cdot \text{ eqn 2}\]
From \[\text{eqn 1}\] and \[\text{eqn 2}\] , we can write,
 \[\dfrac{\left( x+y \right)q-yr}{x}=\dfrac{\left( x-y \right)q+xr}{y}\]
 \[\Rightarrow yxq+{{y}^{2}}q-{{y}^{2}}r={{x}^{2}}q-xyq+{{x}^{2}}r\]
 \[\Rightarrow 2yxq+\left( {{y}^{2}}-{{x}^{2}} \right)q=\left( {{x}^{2}}+{{y}^{2}} \right)r\]
 \[\Rightarrow \dfrac{q}{\left( {{x}^{2}}+{{y}^{2}} \right)}=\dfrac{r}{2yx+\left( {{y}^{2}}-{{x}^{2}} \right)}\text{ }\cdot \cdot \cdot \cdot \cdot \cdot \cdot \text{ eqn 3}\]

Now, from equation \[xp=\left( x+y \right)q-yr\] , we can write,
 \[\therefore r=\dfrac{\left( x+y \right)q-xp}{y}\text{ }\cdot \cdot \cdot \cdot \cdot \cdot \cdot \text{ eqn 4}\]
And from equation \[yp=\left( x-y \right)q+xr\] , we can write,
 \[\therefore r=\dfrac{yp-\left( x-y \right)q}{x}\text{ }\cdot \cdot \cdot \cdot \cdot \cdot \cdot \text{ eqn 5}\]
 From \[\text{eqn 4}\] and \[\text{eqn 5}\] , we can write,
 \[\dfrac{\left( x+y \right)q-xp}{y}=\dfrac{yp-\left( x-y \right)q}{x}\]
 \[\Rightarrow {{x}^{2}}q+xyq-{{x}^{2}}p={{y}^{2}}p-xyq+{{y}^{2}}q\]
 \[\Rightarrow \left( {{x}^{2}}-{{y}^{2}}+2xy \right)q=\left( {{x}^{2}}+{{y}^{2}} \right)p\]
 \[\Rightarrow \dfrac{q}{\left( {{x}^{2}}+{{y}^{2}} \right)}=\dfrac{p}{\left( {{x}^{2}}-{{y}^{2}}+2xy \right)}\text{ }\cdot \cdot \cdot \cdot \cdot \cdot \cdot \text{ eqn 6}\]
From \[\text{eqn 3}\] and \[\text{eqn 6}\] , we can write,
 \[\dfrac{p}{\left( {{x}^{2}}-{{y}^{2}}+2xy \right)}=\dfrac{q}{\left( {{x}^{2}}+{{y}^{2}} \right)}=\dfrac{r}{\left( {{y}^{2}}-{{x}^{2}}+2xy \right)}\]
The above equation represents the general expression for three positive integers in Arithmetic Progression, and the sum of every two numbers is a perfect square. Here \[\dfrac{x}{y}\] represents a constant ratio.

Note: In such types of sum, first thing that we have to keep in mind is the general equation for an A.P i.e. If a, b, c are in A.P, then \[a-b=b-c\] . In this sum, what we do is, try to express the equation of the form \[\dfrac{p}{k1}=\dfrac{q}{k2}=\dfrac{r}{k3}\] , thus this evaluates to a constant value and we are done.