Prime p; p^2 is the arithmetic mean of m^2 and n^2

 
 

Let   p   be a prime number and that there are 2 distinct positive integers   m   and   n
such that   p^2   is the arithmetic mean of   m^2   and   n^2

Prove that   2 \, p - m - n   is either a square or twice a square.

 
 

For example, let’s go thru all prime numbers < 100

1/2 \; (m^2 + n^2) = 5^2   ………..   m = \pm 7,     n = \pm 1
1/2 \; (m^2 + n^2) = 13^2   ……….   m = \pm 17,     n = \pm 7
1/2 \; (m^2 + n^2) = 17^2   ……….   m = \pm 23,     n = \pm 7
1/2 \; (m^2 + n^2) = 29^2   ……….   m = \pm 41,     n = \pm 1
1/2 \; (m^2 + n^2) = 37^2   ……….   m = \pm 47,     n = \pm 23
1/2 \; (m^2 + n^2) = 41^2   ……….   m = \pm 49,     n = \pm 31
1/2 \; (m^2 + n^2) = 53^2   ……….   m = \pm 73,     n = \pm 17
1/2 \; (m^2 + n^2) = 61^2   ……….   m = \pm 71,     n = \pm 49
1/2 \; (m^2 + n^2) = 73^2   ……….   m = \pm 103,     n = \pm 7
1/2 \; (m^2 + n^2) = 89^2   ……….   m = \pm 119,     n = \pm 41
1/2 \; (m^2 + n^2) = 97^2   ……….   m = \pm 137,     n = \pm 7

2\times 5 \; - \; 7 \; - \; 1 \; = \; 2 \; = \; 2 \; \times \; 1^2
2\times 13 \; - \; 17 \; - \; 7 \; = \; 2 \; = \; 2 \; \times \; 1^2
2\times 17 \; - \; 23 \; - \; 7 \; = \; 4 \; = \; 2^2
2\times 29 \; - \; 41 \; - \; 1 \; = \; 16 \; = \; 4^2
2\times 37 \; - \; 47 \; - \; 23 \; = \; 4 \; = \; 2^2
2\times 41 \; - \; 49 \; - \; 31 \; = \; 2 \; = \; 2 \; \times \; 1^2
2\times 53 \; - \; 73 \; - \; 17 \; = \; 16 \; = \; 4^2
2\times 61 \; - \; 71 \; - \; 49 \; = \; 2 \; = \; 2 \; \times \; 1^2
2\times 73 \; - \; 103 \; - \; 7 \; = \; 36 \; = \; 6^2
2\times 89 \; - \; 119 \; - \; 41 \; = \; 18 \; = \; 2 \; \times \; 3^2
2\times 97 \; - \; 137 \; - \; 7 \; = \; 50 \; = \; 2 \; \times \; 5^2

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

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About benvitalis

math grad - Interest: Number theory
This entry was posted in Prime Numbers and tagged . Bookmark the permalink.

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