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Every whole number is a sum of four squares: 7 = 4 + 1 + 1 + 1, and 7 cannot be done with fewer. What about cubes, fourth powers or sixth powers? Asking how many you need, and proving the answer, is Waring's problem. One of its hardest cases was settled by the Indian number theorist S. S. Pillai. This post explains the problem, shows why the number 73 appears, and covers what can be verified about Pillai's life.
What is Waring's problem?
Wolfram MathWorld summarises it. In 1770 Edward Waring suggested that for each power k there is a number g(k) such that every positive integer is a sum of at most g(k) k-th powers. Lagrange had proved g(2) = 4, the four-square theorem. David Hilbert proved in 1909 that g(k) exists for every k. Our post on David Hilbert and his problems covers more of his work.
Knowing that g(k) exists is not the same as knowing its value. MathWorld lists the known values, including g(3) = 9, g(4) = 19, g(5) = 37 and g(6) = 73, and notes that Dickson, Pillai and Niven worked on explicit formulas for g(k).
Why does the answer have to be at least 73?
There is a neat lower bound, which MathWorld attributes to J. A. Euler, a son of Leonhard Euler:
g(k) ≥ 2^k + ⌊(3/2)^k⌋ − 2
Here ⌊ ⌋ means round down. The idea is to find a number that is small enough to use only the powers 1^k and 2^k, but awkward to build from them.
Worked example for k = 6.
2⁶ = 64 and (3/2)⁶ = 11.39, which rounds down to 11.
Take n = 64 × 11 − 1 = 703.
The next sixth power is 3⁶ = 729, which is bigger than 703, so only 1 and 64 are available.
Using q copies of 64 leaves 703 − 64q ones, for a total of 703 − 63q terms. This is smallest when q is as large as possible, q = 10: 640 + 63 ones = 703, which is 10 + 63 = 73 terms.
So 703 needs 73 sixth powers, and g(6) cannot be less than 73. The same recipe gives 23 = 8 + 8 + seven 1s for cubes (9 terms) and 79 = four 16s plus fifteen 1s for fourth powers (19 terms).
What did Pillai prove?
Pillai's paper On Waring's problem g(6) = 73 appeared in July 1940 in the Proceedings of the Indian Academy of Sciences, Section A, volume 12. Its title states the result: 73 sixth powers are always enough, so the lower bound is exact. It was one paper in a long run. His other titles include "On Waring's Problem" in the Journal of the Indian Mathematical Society in 1937 and "On Waring's Problem IX" in 1939, showing he returned to the question again and again.
What is Pillai's conjecture?
His name is also on an open problem. Michael Bennett's paper On some exponential equations of S. S. Pillai (2001) explains that in the 1930s and 1940s Pillai studied equations a^x − b^y = c, where c is fixed. He conjectured that for each such c there are only finitely many solutions with a, b, x and y all greater than 1. Bennett calls this "an outstanding open problem". The case c = 1 is Catalan's equation, whose famous solution is 3² − 2³ = 9 − 8 = 1.
What is known about his life?
Less than one would like, and this post sticks to what could be checked. A 2004 article in Resonance by B. Sury, S S Pillai: 5 April 1901–31 August 1950, gives his dates in its title. An obituary in the Journal of the Indian Mathematical Society (1951) records that he died in the crash of the aircraft Star of Maryland near Cairo on 31 August 1950, while on his way to the United States, and calls him one of India's best mathematicians.
Which mistakes does this help you avoid?
Forgetting 1 is a power. 1 = 1^k for every k, and the worked example depends on it.
Rounding the wrong way. ⌊11.39⌋ is 11, not 12. Using 12 gives 767, which is larger than 729 and breaks the argument.
Confusing "some numbers need" with "all numbers need". Most numbers need far fewer than 73 sixth powers. g(6) is about the worst case.
What can a student take from Pillai?
He picked a hard, precise question and kept working on it across many papers. Number theory problems in olympiads reward the same habit: try small cases, find the worst case, then try to prove nothing is worse. Our guide to IOQM number theory questions has practice in that style.
When a proof step stops making sense, the Fermi Problem Library has a step-by-step solver that explains the reasoning rather than printing an answer, so you can check where your own argument went off.
Try it now: use the formula to find the number that needs 37 fifth powers, then check it by hand.
