Integer n such that 2*n^2 can be written as sum of two squares in four ways

 
 
Find the least integer   n   such that   2 \, n^2   can be written as sum of two squares in four ways

 
Also, find the next few ones.
 
 
Find the least integer   m   such that   2 \, m^2   can be written as sum of two squares in five ways

 

a^2 \; ... \; b^2 \; ... \; c^2
d^2 \; ... \; e^2 \; ... \; f^2
g^2 \; ... \; h^2 \; ... \; i^2

a^2 + i^2 \; = \; b^2 + h^2 \; = \; c^2 + g^2 \; = \; f^2 + d^2 \; = \; 2 \, e^2

 
13^2 \; ... \; 23^2 \; ... \; 47^2
35^2 \; ... \; 65^2 \; ... \; 85^2
79^2 \; ... \; 89^2 \; ... \; 91^2

13^2 + 91^2 \; = \; 23^2 + 89^2 \; = \; 47^2 + 79^2 \; = \; 85^2 + 35^2 \; = \; 2 \,(65^2)
 
                                ——————————             
 
Or, we could use the identities:

(a^2 + b^2 + c^2 + d^2)^2 = (a^2 + b^2 - c^2 - d^2)^2 + (2 \,(b \, c - a \, d))^2 + (2 \,(a \, c + b \, d))^2
(a^2 + b^2 + c^2 + d^2)^2 = (2 \,(a \, d + b \, c))^2 + (a^2 - b^2 + c^2 - d^2)^2 + (2 \,(c \, d - a \, b))^2
(a^2 + b^2 + c^2 + d^2)^2 = (2 \,(b \, d - a \, c))^2 + (2 \,(a \, b + c \, d))^2 + (a^2 - b^2 - c^2 + d^2)^2
(a^2 + b^2 + c^2 + d^2)^2 = (a^2 + b^2 - c^2 - d^2)^2 + (2 \,(a \, d + b \, c))^2 + (2 \,(b \, d - a \, c))^2
(a^2 + b^2 + c^2 + d^2)^2 = (2 \,(b \, c - a \, d))^2 + (a^2 - b^2 + c^2 - d^2)^2 + (2 \,(a \, b + c \, d))^2
(a^2 + b^2 + c^2 + d^2)^2 = (2 \,(a \, c + b \, d))^2 + (2 \,(c \, d - a \, b))^2 + (a^2 - b^2 - c^2 + d^2)^2
 
to form a 3 x 3 grid:
 

( \,a^2 + b^2 - c^2 - d^2 \,)^2   ……….   ( \,2 \,(b \,c - a \,d) \,)^2   …………………   ( \,2 \,(a \,c + b \,d) \,)^2

( \,2 \,(a \,d + b \,c) \,)^2   …………….   ( \,a^2 - b^2 + c^2 - d^2 \,)^2   ……………   ( \,2 \,(c \,d - a \,b) \,)^2

( \,2 \,(b \,d - a \,c) \,)^2   …………………   ( \,2 \,(a \,b + c \,d) \,)^2   …………….   ( \,a^2 - b^2 - c^2 + d^2 \,)^2

 
                                ——————————             

–>   there’s an infinite family containing   1^2   :

(10 \, n^2 + 16 \, n + 9)^2
= \; 1^2 \; + \; (8+10 \,n)^2 \; + \; (4+16 \,n+10 \,n^2)^2
= \; (4+8 \,n+6 \,n^2)^2 \; + \; (4+14 \,n+8 \,n^2)^2 \; + \; (7+8 \,n)^2
= \; (8+14 \,n+8 \,n^2)^2 \; + \; (1+8 \,n+6 \,n^2)^2 \; + \; (4+6 \,n)^2
= \; 1^2 \; + \; (4+8 \,n+6 \,n^2)^2 \; + \; (8+14 \,n+8 \,n^2)^2
= \; (8+10 \,n)^2 \; + \; (4+14 \,n+8 \,n^2)^2 \; + \; (1+8 \,n+6 \,n^2)^2
= \; (4+16 \,n+10 \,n^2)^2 \; + \; (7+8 \,n)^2 \; + \; (4+6 \,n)^2

 

1^2   ……………………………..   ( \,8+10 \,n \,)^2   ………………..   ( \,4+16 \,n+10 \,n^2 \,)^2

( \,4+8 \,n+6n^2 \,)^2   ……….   ( \,4+14 \,n+8 \,n^2 \,)^2   ……….   ( \,7+8 \,n \,)^2

( \,8+14 \,n+8 \,n^2 \,)^2   ………   ( \,1+8 \,n+6 \,n^2 \,)^2   ………..   ( \,4+6 \,n \,)^2

 

Similarly an infinite family containing   2^2   is given by

(5 \, n^2 + 14 \, n + 15)^2
= \; 2^2 \; + \; (14+10 \,n)^2 \; + \; (5+14 \,n+5 \,n^2)^2
= \; (11+10 \,n+3 \,n^2)^2 \; + \; (2+10 \,n+4 \,n^2)^2 \; + \; (10+8 \,n)^2
= \; (10+10 \,n+4 \,n^2)^2 \; + \; (5+10 \,n+3 \,n^2)^2 \; + \; (10+6 \,n)^2
= \; 2^2 \; + \; (11+10 \,n+3 \,n^2)^2 \; + \; (10+10 \,n+4 \,n^2)^2
= \; (14+10 \,n)^2 \; + \; ( \,2+10 \,n+4 \,n^2 \,)^2 \; + \; (5+10 \,n+3 \,n^2)^2
= \; (5+14 \,n+5 \,n^2)^2 \; + \; (10+8 \,n)^2 \; + \; (10+6 \,n)^2

2^2   …………………………………………..   ( \,14+10 \,n \,)^2   ………………..   ( \,5+14 \,n+5 \,n^2 \,)^2

( \,11+10 \,n+3 \,n^2 \,)^2   ……………   ( \,2+10 \,n+4 \,n^2 \,)^2   …………….   ( \,10+8 \,n \,)^2

( \,10+10 \,n+4 \,n^2 \,)^2   ……………   ( \,5+10 \,n+3 \,n^2 \,)^2   …………….   ( \,10+6 \,n \,)^2

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

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About benvitalis

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

3 Responses to Integer n such that 2*n^2 can be written as sum of two squares in four ways

  1. Paul says:

    The least integer is 65, the next few are:- ,85, 130, 145, 170.

    and 5 terms is

    725
    P

  2. benvitalis says:

    Other possible solutions

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