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Let $\mathbb{N}$ be the set of positive integers.

The function $\sigma(N)$ gives the sum of the divisors of $N$.

My question is:

Does the following equation have any solutions for $x \in \mathbb{N}$? $$\sigma(3x + 1) = 4x + 1$$

Notice that $3x + 1$ must be deficient, and that if we allow $x = 0$, then it is a (trivial) solution.

Furthermore, suppose that $3x + 1 = p^{\alpha}$, where $p$ is prime. Then we have $$\sigma(3x + 1) = \frac{p^{\alpha+1} - 1}{p - 1} = \frac{4(p^{\alpha} - 1)}{3} + 1 = 4x + 1,$$ from which we obtain $$3(p^{\alpha+1} - 1) = 4(p - 1)(p^{\alpha} - 1) + 3(p - 1).$$ Simplifying and collecting like terms, we get $$0 = p^{\alpha+1} - 4p^{\alpha} + p + 4.$$ Rewriting the last equation, we have $$(p^{\alpha} + 1)(4 - p) = 8.$$ As $p^{\alpha} + 1 \geq 3$, we have the possibilities $$(p^{\alpha} + 1, 4 - p) \in \{(4,2),(8,1)\}.$$ Both of them are impossible under the given constraints.

Consequently, $\omega(3x + 1) \geq 2$, where $\omega(y)$ is the number of distinct prime factors of $y$. (That is, $3x + 1$ must be composite.)

Lastly, I tried checking for equality (in the range $0 \leq x \leq 100$) using a spreadsheet:

A B C D E

0 1 1 1 YES

1 4 7 5 NO

2 7 8 9 NO

3 10 18 13 NO

4 13 14 17 NO

5 16 31 21 NO

6 19 20 25 NO

7 22 36 29 NO

8 25 31 33 NO

9 28 56 37 NO

10 31 32 41 NO

11 34 54 45 NO

12 37 38 49 NO

13 40 90 53 NO

14 43 44 57 NO

15 46 72 61 NO

16 49 57 65 NO

17 52 98 69 NO

18 55 72 73 NO

19 58 90 77 NO

20 61 62 81 NO

21 64 127 85 NO

22 67 68 89 NO

23 70 144 93 NO

24 73 74 97 NO

25 76 140 101 NO

26 79 80 105 NO

27 82 126 109 NO

28 85 108 113 NO

29 88 180 117 NO

30 91 112 121 NO

31 94 144 125 NO

32 97 98 129 NO

33 100 217 133 NO

34 103 104 137 NO

35 106 162 141 NO

36 109 110 145 NO

37 112 248 149 NO

38 115 144 153 NO

39 118 180 157 NO

40 121 133 161 NO

41 124 224 165 NO

42 127 128 169 NO

43 130 252 173 NO

44 133 160 177 NO

45 136 270 181 NO

46 139 140 185 NO

47 142 216 189 NO

48 145 180 193 NO

49 148 266 197 NO

50 151 152 201 NO

51 154 288 205 NO

52 157 158 209 NO

53 160 378 213 NO

54 163 164 217 NO

55 166 252 221 NO

56 169 183 225 NO

57 172 308 229 NO

58 175 248 233 NO

59 178 270 237 NO

60 181 182 241 NO

61 184 360 245 NO

62 187 216 249 NO

63 190 360 253 NO

64 193 194 257 NO

65 196 399 261 NO

66 199 200 265 NO

67 202 306 269 NO

68 205 252 273 NO

69 208 434 277 NO

70 211 212 281 NO

71 214 324 285 NO

72 217 256 289 NO

73 220 504 293 NO

74 223 224 297 NO

75 226 342 301 NO

76 229 230 305 NO

77 232 450 309 NO

78 235 288 313 NO

79 238 432 317 NO

80 241 242 321 NO

81 244 434 325 NO

82 247 280 329 NO

83 250 468 333 NO

84 253 288 337 NO

85 256 511 341 NO

86 259 304 345 NO

87 262 396 349 NO

88 265 324 353 NO

89 268 476 357 NO

90 271 272 361 NO

91 274 414 365 NO

92 277 278 369 NO

93 280 720 373 NO

94 283 284 377 NO

95 286 504 381 NO

96 289 307 385 NO

97 292 518 389 NO

98 295 360 393 NO

99 298 450 397 NO

100 301 352 401 NO

Column A = $x$

Column B = $3x + 1$

Column C = $\sigma(3x + 1)$

Column D = $4x + 1$

Column E = Test whether Column C equals Column D

I am therefore compelled to predict that:

The equation $\sigma(3x + 1) = 4x + 1$ does not have any solutions for $x \in \mathbb{N}$.

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  • 1
    $\begingroup$ Maybe you can utilize a formula form the following link:math.stackexchange.com/questions/22721/… $\endgroup$ – NoChance Oct 13 '15 at 8:17
  • $\begingroup$ Thank you very much for pointing me to that MSE link, @NoChance! $\endgroup$ – Jose Arnaldo Bebita-Dris Oct 13 '15 at 8:23
  • $\begingroup$ Some small results I've got are (1) $2\not\mid 3x+1$ (2) $3\not\mid 3x+1$ (3) $3x+1$ has to be a perfect square. $\endgroup$ – mathlove Nov 6 '18 at 11:07
  • $\begingroup$ I would be more than delighted to hear about your results, @mathlove. =) $\endgroup$ – Jose Arnaldo Bebita-Dris Nov 6 '18 at 11:31
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Claim 1 : $2\not\mid 3x+1$.

Proof : $x=1$ is not a solution.

For $x\ge 2$, suppose that $2$ is a divisor of $3x+1$. Then, we have $$4x+1=\sigma(3x+1)\ge 1+(3x+1)+2+\frac{3x+1}{2}$$ There are no $x\ge 2$ satisfying the inequality.$\quad\square$

Claim 2 : $3x+1$ is a perfect square.

Proof : Let $3x+1={p_1}^{a_1}{p_2}^{a_2}\cdots {p_k}^{a_k}$ where $p_1,p_2,\cdots, p_k$ are distinct primes larger than $3$ and $a_i\ge 1\in\mathbb Z$. Then, the equation becomes $$(1+\cdots +{p_1}^{a_1})(1+\cdots +{p_2}^{a_2})\cdots (1+\cdots +{p_k}^{a_k})=4x+1$$The RHS is odd since $x$ is even. So, $a_i$ has to be even for every $i$.$\quad\square$

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  • $\begingroup$ Just a minor comment: Surely $3 \not\mid 3x+1$ as $3x+1 \equiv 1 \pmod 3$? =) $\endgroup$ – Jose Arnaldo Bebita-Dris Nov 6 '18 at 11:42
  • $\begingroup$ @Jose Arnaldo Bebita Dris : You are right:) Deleting it. $\endgroup$ – mathlove Nov 6 '18 at 11:43

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