## 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

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

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

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

$13^2 + 91^2 \; = \; 23^2 + 89^2 \; = \; 47^2 + 79^2 \; = \; 85^2 + 35^2 \; = \; 2 \,(65^2)$

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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:

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

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

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

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–>   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

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

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

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

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

math grad - Interest: Number theory
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### 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

• benvitalis says:

Nice.
And we can place them in a 3 x 3 grid

2. benvitalis says:

Other possible solutions