Intra numeral mixing with Pythagorean triples

A most effective and quick way of getting R^2-relation

2 2

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

= 101(a^2+b^2)+20(ac+bd)

or

(10c+b)^2+(10d+a)^2= (10a+d)^2+(10b+c)^2

= 101(a^2+b^2)+20(ad+bc).

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+nc)^2+(mb+nd)^2= (mc+na)^2+(md+nb)^2

=(m^2+n^2)(a^2+b^2)+2mn(ac+db)

or

(ma+nd)^2+(mb+nc)^2= (mc+nb)^2+ (md+na)^2

= (m^2+n^2)(a^2+b^2)+ 2mn(ad+bc)

A most effective and quick way of getting R^2-relation

2 2

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

= 101(a^2+b^2)+20(ac+bd)

or

(10c+b)^2+(10d+a)^2= (10a+d)^2+(10b+c)^2

= 101(a^2+b^2)+20(ad+bc).

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+nc)^2+(mb+nd)^2= (mc+na)^2+(md+nb)^2

=(m^2+n^2)(a^2+b^2)+2mn(ac+db)

or

(ma+nd)^2+(mb+nc)^2= (mc+nb)^2+ (md+na)^2

= (m^2+n^2)(a^2+b^2)+ 2mn(ad+bc)

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