Diagonals-rectangular cube

Imagine any rectangular cube say with side lengths x,y and z
There will be 3 diagonals -we can call them as a,b and c
It will be very difficult to find values of x,y and z so that the diagonal lengths
a,b and c have integral values.
There is a complicated formula to find such values-
Here are some examples of such values-
Example 1) side lengths 44,117 and 240 .The diagonals have values  125,244 and 267
Example 2) side lengths 352,  936 and 1920.  The diagonals have values 1000,1952 and 2136.
Example 3) side lengths 828,2035 and 3120.The diagonals have values 2197,3228 and 3725.  

Perfect triangles

Triangles for which the perimeter and the area have the same integral values are called
perfect triangles-Some examples below-
1)Sides 6,8,10,-perimeter and area=24
2)Sides 5,12,13-perimeter and area=30
3)Sides 9,10,17-perimeter and area=36
4)Sides 7,15,20-perimeter and area=42
5)Sides 6,25,29-perimeter and area=60
You can see all the integral values are multiples of 6.
For your information the areas can be calculated using the following formula-
Find the perimeter 'p' and divide it by 2 to get 's' the semi-perimeter.
Assuming the sides are designated by a, b and c find the product of s,s-a,s-b and s-c
The square root of the product is the area
Example (4) above......a=7,b=15,c=20.....p=42....s=21
s-a=14,s-b=6,s-c=1.Hence the product is 21*14*6*1=1764...Sq root=42=area

Connection between numbers 39 and 1200

You can split up the number 39 into 3 parts in 3different ways,and in each case the products of the 3 parts will be the same viz 1200 as shown below-
25+8+6=39...25*8*6=1200
24+10+5=39....24*10*5=1200
20+15+4=39....20*15*4=1200

Using odd numbers for squares/cubes

Here are ways for obtaining squares/cubes using odd numbers-
Consider the following set of odd numbers-
1,3,5,7,9,11,13,15,17,19,21,23,25,27,29,31,33,35,37etc
For getting squares you simply add them consecutively-
1=1*1
1+3=4=2*2
1+3+5=9=3*3
1+3+5+7=16=4*4etc
1+3+5+.......37=361=19*19 and further squares.
For getting cubes you add as follows-
1=cube of 1
3+5=8=cube of 2
7+9+11=27=cube of 3
13+15+17+19=64=cube of 4
21+23+25+27+29=125=cube of 5 etc
Considering  the above groups as the 1st,2nd,3rd,4th &5th, the 'n'th group will be as follows-
First term n^2-n+1
Last term n^2+n-1
No of terms-n
Sum of all the terms=n*average of the first &last term=n*n^2=n^3=cube of n
For example the 30th group will have 871 as the first term,929 as the last term and the sum of
all the terms will thus be 30*(871+929)/2=30*900=27000=cube of 30