<h3 data-start="281" data-end="317" class="">🔍 Understanding the Problem</h3>

We are to find three rational square numbers in arithmetic progression, i.e.,

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><msup><mrow><mo fence="true">(</mo><mfrac><mrow><mi>a</mi><mo>−</mo><mi>d</mi></mrow><mi>b</mi></mfrac><mo fence="true">)</mo></mrow><mn>2</mn></msup><mo separator="true">,</mo><msup><mrow><mo fence="true">(</mo><mfrac><mi>a</mi><mi>b</mi></mfrac><mo fence="true">)</mo></mrow><mn>2</mn></msup><mo separator="true">,</mo><msup><mrow><mo fence="true">(</mo><mfrac><mrow><mi>a</mi><mo>+</mo><mi>d</mi></mrow><mi>b</mi></mfrac><mo fence="true">)</mo></mrow><mn>2</mn></msup></mrow><annotation encoding="application/x-tex">\left(\frac{a - d}{b}\right)^2, \left(\frac{a}{b}\right)^2, \left(\frac{a + d}{b}\right)^2</annotation></semantics></math>(ba−d​)2,(ba​)2,(ba+d​)2

with the property that the common difference between their squares is a fixed number $c$, i.e.,

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><msup><mrow><mo fence="true">(</mo><mfrac><mrow><mi>a</mi><mo>+</mo><mi>d</mi></mrow><mi>b</mi></mfrac><mo fence="true">)</mo></mrow><mn>2</mn></msup><mo>−</mo><msup><mrow><mo fence="true">(</mo><mfrac><mi>a</mi><mi>b</mi></mfrac><mo fence="true">)</mo></mrow><mn>2</mn></msup><mo>=</mo><mi>c</mi><mspace linebreak="newline"></mspace><msup><mrow><mo fence="true">(</mo><mfrac><mi>a</mi><mi>b</mi></mfrac><mo fence="true">)</mo></mrow><mn>2</mn></msup><mo>−</mo><msup><mrow><mo fence="true">(</mo><mfrac><mrow><mi>a</mi><mo>−</mo><mi>d</mi></mrow><mi>b</mi></mfrac><mo fence="true">)</mo></mrow><mn>2</mn></msup><mo>=</mo><mi>c</mi></mrow><annotation encoding="application/x-tex">\left(\frac{a + d}{b}\right)^2 - \left(\frac{a}{b}\right)^2 = c \\ \left(\frac{a}{b}\right)^2 - \left(\frac{a - d}{b}\right)^2 = c</annotation></semantics></math>(ba+d​)2−(ba​)2=c(ba​)2−(ba−d​)2=c

These simplify to:

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mrow><mo fence="true">(</mo><mfrac><mrow><mo stretchy="false">(</mo><mi>a</mi><mo>+</mo><mi>d</mi><msup><mo stretchy="false">)</mo><mn>2</mn></msup><mo>−</mo><msup><mi>a</mi><mn>2</mn></msup></mrow><msup><mi>b</mi><mn>2</mn></msup></mfrac><mo fence="true">)</mo></mrow><mo>=</mo><mi>c</mi><mo>⇒</mo><mfrac><mrow><mn>2</mn><mi>a</mi><mi>d</mi><mo>+</mo><msup><mi>d</mi><mn>2</mn></msup></mrow><msup><mi>b</mi><mn>2</mn></msup></mfrac><mo>=</mo><mi>c</mi></mrow><annotation encoding="application/x-tex">\left(\frac{(a + d)^2 - a^2}{b^2}\right) = c \Rightarrow \frac{2ad + d^2}{b^2} = c</annotation></semantics></math>(b2(a+d)2−a2​)=c⇒b22ad+d2​=c

Similarly:

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mfrac><mrow><mn>2</mn><mi>a</mi><mi>d</mi><mo>−</mo><msup><mi>d</mi><mn>2</mn></msup></mrow><msup><mi>b</mi><mn>2</mn></msup></mfrac><mo>=</mo><mi>c</mi></mrow><annotation encoding="application/x-tex">\frac{2ad - d^2}{b^2} = c</annotation></semantics></math>b22ad−d2​=c

So we get:

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mfrac><mrow><mn>2</mn><mi>a</mi><mi>d</mi><mo>±</mo><msup><mi>d</mi><mn>2</mn></msup></mrow><msup><mi>b</mi><mn>2</mn></msup></mfrac><mo>=</mo><mi>c</mi><mo>⇒</mo><mn>2</mn><mi>a</mi><mi>d</mi><mo>=</mo><mi>c</mi><msup><mi>b</mi><mn>2</mn></msup><mo>±</mo><msup><mi>d</mi><mn>2</mn></msup></mrow><annotation encoding="application/x-tex">\frac{2ad \pm d^2}{b^2} = c \Rightarrow 2ad = c b^2 \pm d^2</annotation></semantics></math>b22ad±d2​=c⇒2ad=cb2±d2

But most straightforwardly, we can work with:

Let <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msup><mi>x</mi><mn>2</mn></msup><mo separator="true">,</mo><msup><mi>y</mi><mn>2</mn></msup><mo separator="true">,</mo><msup><mi>z</mi><mn>2</mn></msup></mrow><annotation encoding="application/x-tex">x^2, y^2, z^2</annotation></semantics></math>x2,y2,z2 be the squares in arithmetic progression with difference $c$. Then:

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><msup><mi>y</mi><mn>2</mn></msup><mo>−</mo><msup><mi>x</mi><mn>2</mn></msup><mo>=</mo><msup><mi>z</mi><mn>2</mn></msup><mo>−</mo><msup><mi>y</mi><mn>2</mn></msup><mo>=</mo><mi>c</mi><mo>⇒</mo><msup><mi>z</mi><mn>2</mn></msup><mo>−</mo><mn>2</mn><msup><mi>y</mi><mn>2</mn></msup><mo>+</mo><msup><mi>x</mi><mn>2</mn></msup><mo>=</mo><mn>0</mn></mrow><annotation encoding="application/x-tex">y^2 - x^2 = z^2 - y^2 = c \Rightarrow z^2 - 2y^2 + x^2 = 0</annotation></semantics></math>y2−x2=z2−y2=c⇒z2−2y2+x2=0

So we can find such rational triples by solving:

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><msup><mi>x</mi><mn>2</mn></msup><mo>+</mo><msup><mi>z</mi><mn>2</mn></msup><mo>=</mo><mn>2</mn><msup><mi>y</mi><mn>2</mn></msup><mo separator="true">,</mo><mspace width="1em"></mspace><msup><mi>z</mi><mn>2</mn></msup><mo>−</mo><msup><mi>x</mi><mn>2</mn></msup><mo>=</mo><mn>2</mn><mi>c</mi></mrow><annotation encoding="application/x-tex">x^2 + z^2 = 2y^2, \quad z^2 - x^2 = 2c</annotation></semantics></math>x2+z2=2y2,z2−x2=2c <hr data-start="1283" data-end="1286" class=""> <h3 data-start="1288" data-end="1324" class="">✅ Known solution for gap = 7</h3>

Let’s try to find a minimal solution with gap 7.

Suppose <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msup><mi>x</mi><mn>2</mn></msup><mo separator="true">,</mo><msup><mi>y</mi><mn>2</mn></msup><mo separator="true">,</mo><msup><mi>z</mi><mn>2</mn></msup></mrow><annotation encoding="application/x-tex">x^2, y^2, z^2</annotation></semantics></math>x2,y2,z2 are rational squares with common difference $7$.

We can use a known parametrization of three rational squares in AP:

Let’s suppose:

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mi>x</mi><mo>=</mo><mfrac><mrow><mi>a</mi><mo>−</mo><mi>d</mi></mrow><mi>b</mi></mfrac><mo separator="true">,</mo><mspace width="1em"></mspace><mi>y</mi><mo>=</mo><mfrac><mi>a</mi><mi>b</mi></mfrac><mo separator="true">,</mo><mspace width="1em"></mspace><mi>z</mi><mo>=</mo><mfrac><mrow><mi>a</mi><mo>+</mo><mi>d</mi></mrow><mi>b</mi></mfrac><mo>⇒</mo><msup><mi>x</mi><mn>2</mn></msup><mo separator="true">,</mo><msup><mi>y</mi><mn>2</mn></msup><mo separator="true">,</mo><msup><mi>z</mi><mn>2</mn></msup><mtext> are in AP if </mtext><msup><mi>z</mi><mn>2</mn></msup><mo>−</mo><msup><mi>y</mi><mn>2</mn></msup><mo>=</mo><msup><mi>y</mi><mn>2</mn></msup><mo>−</mo><msup><mi>x</mi><mn>2</mn></msup><mo>=</mo><mi>c</mi></mrow><annotation encoding="application/x-tex">x = \frac{a - d}{b},\quad y = \frac{a}{b},\quad z = \frac{a + d}{b} \Rightarrow x^2, y^2, z^2 \text{ are in AP if } z^2 - y^2 = y^2 - x^2 = c</annotation></semantics></math>x=ba−d​,y=ba​,z=ba+d​⇒x2,y2,z2 are in AP if z2−y2=y2−x2=c

The square values are:

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><msup><mrow><mo fence="true">(</mo><mfrac><mrow><mi>a</mi><mo>−</mo><mi>d</mi></mrow><mi>b</mi></mfrac><mo fence="true">)</mo></mrow><mn>2</mn></msup><mo separator="true">,</mo><msup><mrow><mo fence="true">(</mo><mfrac><mi>a</mi><mi>b</mi></mfrac><mo fence="true">)</mo></mrow><mn>2</mn></msup><mo separator="true">,</mo><msup><mrow><mo fence="true">(</mo><mfrac><mrow><mi>a</mi><mo>+</mo><mi>d</mi></mrow><mi>b</mi></mfrac><mo fence="true">)</mo></mrow><mn>2</mn></msup><mo>⇒</mo><mtext>difference is </mtext><mfrac><mrow><mo stretchy="false">(</mo><mi>a</mi><mo>+</mo><mi>d</mi><msup><mo stretchy="false">)</mo><mn>2</mn></msup><mo>−</mo><msup><mi>a</mi><mn>2</mn></msup></mrow><msup><mi>b</mi><mn>2</mn></msup></mfrac><mo>=</mo><mfrac><mrow><mn>2</mn><mi>a</mi><mi>d</mi><mo>+</mo><msup><mi>d</mi><mn>2</mn></msup></mrow><msup><mi>b</mi><mn>2</mn></msup></mfrac></mrow><annotation encoding="application/x-tex">\left(\frac{a - d}{b}\right)^2, \left(\frac{a}{b}\right)^2, \left(\frac{a + d}{b}\right)^2 \Rightarrow \text{difference is } \frac{(a + d)^2 - a^2}{b^2} = \frac{2ad + d^2}{b^2}</annotation></semantics></math>(ba−d​)2,(ba​)2,(ba+d​)2⇒difference is b2(a+d)2−a2​=b22ad+d2​

So to get common difference 7, we solve:

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mfrac><mrow><mn>2</mn><mi>a</mi><mi>d</mi><mo>+</mo><msup><mi>d</mi><mn>2</mn></msup></mrow><msup><mi>b</mi><mn>2</mn></msup></mfrac><mo>=</mo><mn>7</mn><mo>⇒</mo><mn>2</mn><mi>a</mi><mi>d</mi><mo>+</mo><msup><mi>d</mi><mn>2</mn></msup><mo>=</mo><mn>7</mn><msup><mi>b</mi><mn>2</mn></msup><mo>⇒</mo><mi>d</mi><mo stretchy="false">(</mo><mn>2</mn><mi>a</mi><mo>+</mo><mi>d</mi><mo stretchy="false">)</mo><mo>=</mo><mn>7</mn><msup><mi>b</mi><mn>2</mn></msup></mrow><annotation encoding="application/x-tex">\frac{2ad + d^2}{b^2} = 7 \Rightarrow 2ad + d^2 = 7b^2 \Rightarrow d(2a + d) = 7b^2</annotation></semantics></math>b22ad+d2​=7⇒2ad+d2=7b2⇒d(2a+d)=7b2

Let’s try small integers to solve this Diophantine equation.

Try $d=1$:

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mn>1</mn><mo stretchy="false">(</mo><mn>2</mn><mi>a</mi><mo>+</mo><mn>1</mn><mo stretchy="false">)</mo><mo>=</mo><mn>7</mn><msup><mi>b</mi><mn>2</mn></msup><mo>⇒</mo><mn>2</mn><mi>a</mi><mo>+</mo><mn>1</mn><mo>=</mo><mn>7</mn><msup><mi>b</mi><mn>2</mn></msup><mo>⇒</mo><mn>2</mn><mi>a</mi><mo>=</mo><mn>7</mn><msup><mi>b</mi><mn>2</mn></msup><mo>−</mo><mn>1</mn><mo>⇒</mo><mi>a</mi><mo>=</mo><mfrac><mrow><mn>7</mn><msup><mi>b</mi><mn>2</mn></msup><mo>−</mo><mn>1</mn></mrow><mn>2</mn></mfrac></mrow><annotation encoding="application/x-tex">1(2a + 1) = 7b^2 \Rightarrow 2a + 1 = 7b^2 \Rightarrow 2a = 7b^2 - 1 \Rightarrow a = \frac{7b^2 - 1}{2}</annotation></semantics></math>1(2a+1)=7b2⇒2a+1=7b2⇒2a=7b2−1⇒a=27b2−1​

Try <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>b</mi><mo>=</mo><mn>1</mn></mrow><annotation encoding="application/x-tex">b = 1</annotation></semantics></math>b=

Posted by EmmaRobert on 2025-05-16 02:21:21