A most effective and quick way of getting R^2-relation
is intra numeral mixing with Pythagorean triple. This technique is useful
only for numeral relations, where equal numbers of squares with equal
number of digits are equated. If there is any difference in the number of
squares, then that much 0^2 must be added in the shortage side.
Similarly, if all the root numbers do not have equal number of digits, then
they can be made equal without affecting its
value by affixing sufficient number of zeros in front of them.
For example 3^2 + 4^2 = 5^2 is written as
3^2 + 4^2 = 5^2 + 0^2
Intra numeral mixing means the number in one side, is simply
affixed in the front or prefixed at the end of the number in the other side.
In intra numeral mixing pairing can be interchanged, but pairing of
numbers must be same for both sides
For the given example,
Suffixing : 35^2 + 40^2 = 53^2 + 4^2 = 2825
Prefixing : 30^2 + 45^2 = 54^2 + 3^2 = 2925
By repeating the procedure one can generate more and more relations.
35^2 +40^2 = 53^2 + 04^2 gives
5335^2+440^2 = 3553^2 + 4004^2
435^2+ 5340^2 = 4053^2 + 3504^2
30^2+45^2= 54^2+ 03^2 gives,
5430^2+ 345^2= 3054^2 +4503^2
330^2+5445^2= 3054^2+ 4503^2
If we consider the Pythagorean triple (5,12,13), the number of digits
in the root numbers is not equal. Hence before making intra numeral mixing,
it is arranged as
05^2 + 12^2 = 13^2 + 00^2
It gives the following numeral relations,
1305^2+12^2= 513^2+1200^2 = 1703169
1312^2+5^2 = 500^2+1213^2 = 1721369
The mathematics behind it gives more insight on such relations.
The intra numeral mixing made on a^2+b^2=c^2+d^2
(for Pythagorean triple d=0) gives,
(10c+a)^2+(10d+b)^2= (10a+c)^2+ (10b+d)^2
It is noted that in the intra numeral mixing, the numbers in the same
side cannot be paired up, as they do not preserve the balanced condition
of the relation.
In this intra numeral mixing, we simply add with each number 10 times
the paired root number. It is found that the balance
of the new relations is not affected when the mixing is done with different
proportions of the two paired numbers.
For a^2+b^2=c^2+d^2, we have
(ma+nd)^2+(mb+nc)^2= (mc+nb)^2+ (md+na)^2
= (m^2+n^2)(a^2+b^2)+ 2mn(ad+bc)