Sums of consecutive integers that result in a 4-th power

 

1
2 3
4 5 6
7 8 9 10
11 12 13 14 15
16 17 18 19 20 21
22 23 24 25 26 27 28
29 30 31 32 33 34 35 36
37 38 39 40 41 42 43 44 45
46 47 48 49 50 51 52 53 54 55
56 57 58 59 60 61 62 63 64 65 66
67 68 69 70 71 72 73 74 75 76 77 78
79 80 81 82 83 84 85 86 87 88 89 90 91
92 93 94 95 96 97 98 99 100 101 102 103 104 105

………………………………………
……………………………………….

 

1
1+4+5+6=16=2^4
1+4+5+6+11+12+13+14+15 = 3^4
1+4+5+6+11+12+13+14+15+22+23+24+25+26+27+28=4^4

1+4+5+6+11+12+13+14+15+22+23+24+25+26+27+28+37+38+39+40+41+42+43+44+45=5^4

5^4 + (56+57+58+59+60+61+62+63+64+65+66) = 6^4
6^4 + (79+80+81+82+83+84+85+86+87+88+89+90+91) = 7^4

 

Show that adding all the numbers that are not crossed off from 1 to the end of a line will give us a 4-th power.

 
 
 

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

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About benvitalis

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

4 Responses to Sums of consecutive integers that result in a 4-th power

  1. osedax says:

    I believe you are missing 47, 48, & 49 for this to work.

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