3 x 3 magic square w/ entries are distinct integer squares

 
 
       A^2   …..   B^2   …..   C^2
       D^2   …..   E^2   …..   F^2
       G^2   …..   H^2   …..   I^2

is a magic square,
 

Can you find a number   m   such that

A^2 + B^2 + C^2
= \; D^2 + E^2 + F^2
= \; G^2 + H^2 + I^2
= \; A^2 + D^2 + G^2
= \; B^2 + E^2 + H^2
= \; C^2 + F^2 + I^2
= \; A^2 + E^2 + I^2
= \; C^2 + E^2 + G^2 \; =  \; m

                               ——————————————–             

A \; ..... \; B \; ..... \; C
D \; ..... \; E \; ..... \; F
G \; ..... \; H \; ..... \; I

A+B+C = 3 \,E \; ..... \; D+E+F = 3 \,E \; ..... \; G+H+I = 3 \,E \; ..... \; A+E+I = 3 \,E
A+D+G = 3 \,E \; ..... \; B+E+H = 3 \,E \; ..... \; C+F+I = 3 \,E \; ..... \; C+E+G = 3 \,E

A+I \; = \; B+H \; = \; C+G \; = \; D+F \; = \; 2 \,E

A+I \; = \; 2 \,E
(A-E) \; + \; (I-E) \; = \; 0
(A-E) \; = \; (E-I)

If we define   n = A-E   and   m = C-E   the magic square contains the terms

E+n \; .......... \; E+m
                 E
E-m \; .......... \; E-n

 
From this the remaining terms follow,   due to the requirement that each row and column sum to   3 \,E

E+n                  E-n-m                  E+m
E-n+m                  E                          E+n-m
E-m                  E+n+m                  E-n

 
Imagine a magic square comprised entirely of square integers.

a^2     b^2     c^2
d^2     e^2     f^2
g^2     h^2     i^2

a^2 = E+n              b^2 = E-n-m             c^2 = E+m
d^2 = E-n+m              e^2 = E                     f^2 = E+n-m
g^2 = E-m              h^2 = E+n+m             i^2 = E-n

 

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

 
a^2 \, i^2 \; = \; (E+n) \,(E-n) \; = \; E^2 \; - \; n^2
a^2 \, i^2 \; + \; n^2 \; = \; E^2 \; = \; e^4

b^2 \, h^2 \; = \; (E-n-m) \,(E+n+m) \; = \; E^2 \; - \; (n+m)^2
b^2 \, h^2 \; + \; (n+m)^2 \; = \; E^2 \; = \; e^4

c^2 \, g^2 \; = \; (E+m) \,(E-m) \; = \; E^2 \; - \; m^2
c^2 \, g^2 \; + \; m^2 \; = \; E^2 \; = \; e^4

f^2 \, d^2 \; = \; (E+n-m) \,(E-n+m) \; = \; E^2 \; - \; (n-m)^2
f^2 \, d^2 \; + \; (n-m)^2 \; = \; E^2 \; = \; e^4

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

Posted in Number Puzzles | Tagged | 3 Comments

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

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

Posted in Number Puzzles | Tagged | 3 Comments

4X4 grid – gcd from 1 to 24

 
 
To place distinct integers in a   n × n   grid such that all gcds from   1   to   2*n(n-1)   are generated.
 

If   n = 4,   then find 16 integers such that all gcds from   1 to   2*4(4-1) = 24 are generated
 
grid-gcd-2

 
Determine the integers so that the sum   A+B+C+…+N+O+P   is minimal
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

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3X3 grid – gcd from 1 to 12

 
 
To place distinct integers in a n × n grid such that all gcds from 1 to 2*n(n-1) are generated.
 

If   n = 3,   then find 9 integers such that all gcds from   1 to   2*3(3-1) = 12 are generated

 
grid-gcd-1

 
Determine the 9 integers so that the sum N is minimal
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

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Integers (a,b,c,d); each pairwise sum & sum of all four is a square

 
 
Find four integers   (a, \; b, \; c, \; d)   such that

a \; + \; b
a \; + \; c
a \; + \; d
b \; + \; c
b \; + \; d
c \; + \; d
a \; + \; b \; + \; c \; + \; d
 
are all squares

 

The first few solutions are:

\{ \, 386, \; 2114, \; 3970, \; 10430 \, \}
\{ \, 590, \; 4594, \; 5810, \; 17906 \, \}
\{ \, 617, \; 15008, \; 26608, \; 63392 \, \}
\{ \, 872, \; 2377, \; 9944, \; 21032 \, \}
\{ \, 2248, \; 4808, \; 12881, \; 22088 \, \}

Also found by pipo

 
\{ \, 386, \; 2114, \; 3970, \; 10430 \, \}

386 + 2114 = 50^2
386 + 3970 = 66^2
386 + 10430 = 104^2
2114 + 3970 = 78^2
2114 + 10430 = 112^2
3970 + 10430 = 120^2
386 + 2114 + 3970 + 10430 = 130^2

\{ \, 590, \; 4594, \; 5810, \; 17906 \, \}

590 + 4594 = 72^2
590 + 5810 = 80^2
590 + 17906 = 136^2
4594 + 5810 = 102^2
4594 + 17906 = 150^2
5810 + 17906 = 154^2
590 + 4594 + 5810 + 17906 = 170^2

\{ \, 617, \; 15008, \; 26608, \; 63392 \, \}

617 + 15008 = 125^2
617 + 26608 = 165^2
617 + 63392 = 253^2
15008 + 26608 = 204^2
15008 + 63392 = 280^2
26608 + 63392 = 300^2
617 + 15008 + 26608 + 63392 = 325^2

\{ \, 872, \; 2377, \; 9944, \; 21032 \, \}

872 + 2377 = 57^2
872 + 9944 = 104^2
872 + 21032 = 148^2
2377 + 9944 = 111^2
2377 + 21032 = 153^2
9944 + 21032 = 176^2
872 + 2377 + 9944 + 21032 = 185^2

\{ \, 2248, \; 4808, \; 12881, \; 22088 \, \}

2248 + 4808 = 84^2
2248 + 12881 = 123^2
2248 + 22088 = 156^2
4808 + 12881 = 133^2
4808 + 22088 = 164^2
12881 + 22088 = 187^2
2248 + 4808 + 12881 + 22088 = 205^2

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

Posted in Number Puzzles | Tagged | 2 Comments

Pythagorean triangle {a+b+c, a^2+b+c, a+(b+c)^2} are to made squares

 
 
Find a Pythagorean triangle   (a, \; b, \; c)   such that
 
a \; + \; b \; + \; c
a^2 \; + \; b \; + \; c
b^2 \; + \; c \; + \; a
c^2 \; + \; a \; + \; b
a \; + \; (b+c)^2
b \; + \; (c+a)^2
c \; + \; (a+b)^2
 
are all squares

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

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Marrakech climate summit COP22 — Palindromic numbers

 
 
climate-summit-1

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

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Diophantine equation : y^2 – p = 2^n, where p is a prime number

 
 
The diophantine equation     y^2 \; - \; p \; = \; 2^{n}

where   p   is a prime number   < 100
 

The only solutions to the diophantine equation     y^2 - 17 = 2^{n}
with   y > 0   are given by     (n, y)   =   (3, 5),   (5, 7),   (6, 9),   (9, 23)

The only solutions to the diophantine equation     y^2 - 41 = 2^{n}
with   y > 0   are given by     (n, y)   =   (3, 7),   (7, 13)

The only solutions to the diophantine equation     y^2 - 73 = 2^{n}
with   y > 0   are given by     (n, y)   =   (3, 9)

The only solutions to the diophantine equation     y^2 - 89 = 2^{n}
with   y > 0   are given by     (n, y)   =   (5, 11),   (13, 91)

The only solutions to the diophantine equation     y^2 - 97 = 2^{n}
with   y > 0   are given by     (n, y)   =   (7, 15)

 
Find solutions for   100 \; < \; p \; < \; 1000
 
 

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

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Diophantine equation : 2^a + 2^b + 2^c = x^2

 
 
Consider the Diophantine equation :

2^{a} \; + \; 2^{b} \; + \; 2^{c} \; = \; x^2

for   a, \; b, \; c, \; x   nonnegative integers,   a \; \leq \; b \; \leq \; c

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

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Diophantine equation : (a-b)/(a+b) + (b-c)/(b+c) + (c-d)/(c+d) + (d-a)/(d+a) = 0

 
 
Determine integral solutions of the Diophantine equation
 
diophantine-1

 
where   a, \; b, \; c, \; d   are distinct
 
Solutions:

b \; \neq \; 0,     d \; = \; (a \, c)/b,     a^2 \, b \, c+a^2 \, c^2+a  \,b^2 \, c+a \, b \, c^2 \; \neq \; 0

If   b = 1,   then     d \; = \; a \, c,     a^2 \, c + a^2 \, c^2 + a \, c + a  \,c^2 \; \neq \; 0

If   b = 2,   then     d = (a \, c)/2,     a^2 \, c^2 + 2 \, a^2 \, c + 2 \, a \, c^2 + 4 \, a \, c \; \neq \; 0
 
 
 
 
 

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

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