Problems & Puzzles: Puzzles

 Puzzle 624. Nice integer quotients Recently I found (*) the following two nice expressions: 2= (2^2+4^2)/(1^2+3^2) 2= (3^2+4^2+...+9^2)/(1^2+2^2+...+7^2) In the first case the overall quotient uses exclusively the square of consecutive integers. No repetition allowed. In the second case, each part of the quotient uses exclusively the square of consecutive integers. You are able to use some of the same integers in both parts. No repetition allowed in the same part. In both cases: -the squares are always added. -the quantity of summands is the same for both parts of the quotient (2 for the first case, 7 for the second case) -the quotient value is the integer 2. Q1. Perhaps you would be interested in getting other expressions alike, for the same quotient value (2) and the same power (2). Q2. Redo Q1 using consecutive prime numbers. _________ (*) See p. 91, Excursions in Number Theory, C. Stanley Ogilvy & J. T. Anderson.

Contributions came from Hakan Summakoğlu, Emmanuel Vantieghem & Farideh Firoozbakht & Jahangeer John Kholdi.

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Hakan wrote:

Q1:
My expressions with smallest n consecutive numbers (for the first case).
N=4:  2= (2^2+4^2)/(1^2+3^2)
N=10:  2= (4^2+7^2+10^2+11^2+12^2)/(3^2+5^2+6^2+8^2+9^2)
N=14:  2= (2^2+6^2+10^2+11^2+12^2+14^2+15^2)/(3^2+4^2+5^2+7^2+8^2+9^2+13^2)
N=18:  2=(1^2+3^2+5^2+9^2+14^2+15^2+16^2+17^2+18^2)/(2^2+4^2+6^2+7^2+8^2+10^2+11^2+12^2+13^2)
N=22:  2=(1^2+2^2+3^2+5^2+15^2+16^2+18^2+19^2+20^2+21^2+22^2)/(4^2+6^2+7^2+8^2+9^2+10^2+11^2+
12^2+13^2+14^2+17^2)
N=26:  2=(1^2+2^2+3^2+4^2+7^2+18^2+20^2+21^2+22^2+23^2+24^2+25^2+26^2)/(5^2+6^2+8^2+9^2+10^2+11^2+
12^2+13^2+14^2+15^2+16^2+17^2+19^2)
N=28:  2=(3^2+4^2+5^2+6^2+17^2+20^2+21^2+23^2+25^2+26^2+27^2+28^2+29^2+30^2)/(7^2+8^2+9^2+10^2+11^2+
12^2+13^2+14^2+15^2+16^2+18^2+19^2+22^2+24^2)
N=32:  2=(2^2+3^2+4^2+5^2+6^2+20^2+21^2+24^2+25^2+26^2+27^2+29^2+30^2+31^2+32^2+33^2)/(7^2+8^2+9^2+10^2+
11^2+12^2+13^2+14^2+15^2+16^2+17^2+18^2+19^2+22^2+23^2+28^2)
N=36: 2=(1^2+2^2+3^2+4^2+5^2+6^2+19^2+20^2+26^2+28^2+29^2+30^2+31^2+32^2+33^2+34^2+35^2+36^2)/(7^2+8^2+9^2+10^2+
11^2+12^2+13^2+14^2+15^2+16^2+17^2+18^2+21^2+22^2+23^2+24^2+25^2+27^2)
N=40:  2=(1^2+2^2+3^2+4^2+5^2+6^2+8^2+22^2+28^2+29^2+30^2+31^2+32^2+34^2+35^2+36^2+37^2+38^2+39^2+40^2)/(7^2+9^2+10^2+
11^2+12^2+13^2+14^2+15^2+16^2+17^2+18^2+19^2+20^2+21^2+23^2+24^2+25^2+26^2+27^2+33^2)
My expressions with smallest n consecutive numbers parts (for the second case).
N=5:  2= (8^2+9^2+...+12^2)/(5^2+6^2+...+9^2)
N=7:  2= (3^2+4^2+...+9^2)/(1^2+2^2+...+7^2)
N=13:  2= (10^2+11^2+...+22^2)/(5^2+6^2+...+17^2)
N=47:  2= (49^2+50^2+...+95^2)/(27^2+28^2+...+73^2)
N=49:  2= (76^2+77^2+...+124^2)/(46^2+47^2+...+94^2)
N=55:  2= (51^2+52^2+...+105^2)/(27^2+28^2+...+81^2)
N=83:  2= (27^2+28^2+...+109^2)/(4^2+5^2+...+86^2)
N=95:  2= (53^2+54^2+...+147^2)/(21^2+22^2+...+115^2)
N=97:  2= (36^2+37^2+...+132^2)/(8^2+9^2+...+104^2)

Q2:
My expressions with smallest n consecutive prime numbers (for the first case).
N=4:  2= (3^2+7^2)/(2^2+5^2)
N=16:  2= (5^2+11^2+19^2+29^2+41^2+43^2+53^2+59^2)/(3^2+7^2+13^2+17^2+23^2+31^2+37^2+47^2)
N=22:  2=(2^2+3^2+5^2+7^2+29^2+47^2+59^2+67^2+71^2+73^2+79^2)/(11^2+13^2+17^2+19^2+23^2+31^2+37^2+41^2+43^2+53^2+61^2)

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Emmanuel wrote:

Q1, first case : I found 16 solutions involving the squares of the first 18 numbers, 243 solutions involving the squares of the first 22 numbers and 1748 solutions involving the squares of the first 26 numbers.  The 'last' found solution was this one :

Numerator = 4134= sum of the squares of  10, 11, 13, 14, 15, 16, 17, 18, 19, 20, 21, 24, 26 ;

Denominator=2067=sum of the squares of  1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 22, 23, 25.

I think that there are many solutions with more numbers.

Second case :  I found two solutions with 5 squares, two with 7 squares, 3 with 13 squares, two with 47 squares and one with 49 squares.  The last one is :

2 = (76^2+77^2+...+124^2)/(46^2+47^2+...+97^2).

Here I also think that there are many other solutions.

Q2, first case.  With the first four primes there is one solution : 2=(3^2+7^2)/(2^2+5^2).

I found 25 solutions involving the first 22 primes, the 'last' of which is this one :

Numerator = 27718=sum of the squares of  2, 3, 19, 29, 41, 43, 59, 61, 67, 71, 79

Denominator = 13859=sum of the squares of  5, 7, 11, 13, 17, 23, 31, 37, 47, 53, 73.

There are also solutions with consecutive primes, not starting with 2.

Second case : I could not find solutions.

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Farideh & Jahangeer wrote:

2 = (5^2+6^2+7^2)/(1^2+2^2+3^2+4^2+5^2)
2 = (15^2+16^2+17^2)/(1^2+2^2+...+10^2)
2 = (21^2+22^2+23^2+24^2)/(1^2+2^2+...+14^2)
2 = (25^2+26^2+27^2)/(1^2+2^2+...+14^2)

Also we have:

2 = (7^3+8^3+...+18^3)/(1^3+2^3+...+15^3)

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