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

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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.

math grad - Interest: Number theory
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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.

• benvitalis says:

Oops. You’re right. I forgot to type those numbers

2. I don’t know what you mean by show, but a simple formula for the sequence would be:

http://www5b.wolframalpha.com/Calculate/MSP/MSP15791ihdf180a2086h42000054i58e1daidi7cge?MSPStoreType=image/gif&s=58

• benvitalis says:

It’s $\sum_{n=1}^{n} \; (2 \, n-1) \, (2 \, (n-1) \, n+1) \; = \; n^4$