Is 7 a congruent number? I have a problem with understanding congruence of number 7. 
The idea is such as: Which integers can occur as the common
difference between three rational squares that form an arithmetic
sequence?
That integer number is congruent
In book I'm reading that : To prove 7 is congruent, observe that $\frac{24}{5}^2,\frac{35}{12}^2, \frac{337}{60}^2$ have common difference 7
But it dosen't work. 
For instance:
$$\frac{337}{60}^2 - \frac{24}{5}^2 = 8.506944444$$
I will be grateful for explaining Best regards
 A: The number $\;7\;$ is congruent since
$$\frac{\frac{24}5\cdot\frac{35}{12}}{2}=7\;,\;\;\text{ and}\;\;\frac{24}5\,,\;\;\frac{35}{12}\;,\;\;\frac{337}{60}\;$$
are the sides of a right triangle with legs' lengths the first two numbers above, and hypotenuse of length the last number.
Adding the info gathered in the comments: putting
$$a=\frac{24}5\;,\;\;b=\frac{35}{12}\;,\;\;c=\frac{337}{60}$$
we then must consider
$$A:=\frac{a-b}2=\frac{113}{120}\;,\;\;B:=\frac c2=\frac{337}{120}\;,\;\;C:=\frac{a+b}2=\frac{463}{120}$$
and now:
$$B^2-A^2=\frac{113,569-12,769}{14,400}=7=\frac{214,369-113,569}{14,400}=C^2-B^2$$
A: I'll make an answer out of my comment:  The statement of the question is in error.  The construction is that if $a^2+b^2 = c^2$ and $n = ab/2$, (so that $n$ is a congruent number) then one can construct an AP of squares via $(a-b)/2$, $c/2$, $(a+b)/2$.  So in this case $(a-b)/2 = 113/120$, $c/2 = 337/120$ and $(a+b)/2 = 463/120$.  Then $$\left({113\over 120}\right)^2 -\left({337\over 120}\right)^2 = 7$$ and $$ \left({463\over 120}\right)^2 -\left({113\over 120}\right)^2= 7$$
A: In general, $n$ is a congruent number if and only if there exists a rational number $x$ such that $(x,x-n,x+n)$ is a triple of squares of rational numbers. 
So $x-n$, $x$ and $x+n$ have common difference $n$, not the sides of the triangle.
The correspondence to rational right triangles $(X,Y,Z)$ with $X<Y<Z$ then is
$$
(X,Y,Z)\mapsto x=(Z/2)^2,
$$
and
$$
x\mapsto (X,Y,Z)=(\sqrt{x+n}-\sqrt{x-n},\sqrt{x+n}+\sqrt{x-n},2\sqrt{x}).
$$
Conrad's notes are beautifully written and have all the details, e.g., section $3$. 
For $n=7$ we obtain the triangle $(\frac{24}5\,,\;\;\frac{35}{12}\;,\;\;\frac{337}{60})$. So we have
$$
(x,x-7,x+7)=\left( \left(\frac{337}{120}\right)^2, \left(\frac{113}{120}\right)^2,\left(\frac{463}{120}\right)^2\right).
$$
