Proof of Absence

There are no distinct natural numbers and that make a square number.

1

If and are coprime, then and are coprime.

2

For two odd numbers and , is even but not divided by 4 while is divided by 4.

3

Sum of two coprime square numbers is not divided by 3 because , , and are not divided by 3.

4

Dividing by the fourth power of the greatest common divisor of and leads to for coprime and . Because is coprime with both and , and must be square numbers.

5

When and are coprime and then and cannot be both odd. Let be even. Then for some coprime and , and from . Then . If and are both odd then cannot be even, so one is odd, the other is even, and .

6

When and are distinct coprime that satisfies with minimal , let be odd and be even. Then and for some coprime and . From the prior equation, is odd and is even. Then from the latter equation, , , and for some coprime , , and with being even. Then leads to . Because is not divided by 3, and are coprime. So and . Then from the prior equation, and for coprime and with one being even and the other being odd. Then , but violates the first minimal condition. Let and are both odd, then and for some coprime and with one being even and the other being odd. Then but there are no such and as seen above.

Reference

Pocklington, Some Diophantine Impossibilities, page 111. Retrieved from Oliver Knill’s homepage.

Proof of Numerosity

There are numerous natural numbers and that make a square number.

Proof of Existence

There are natural numbers and that make a square number.