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Proof of a theorem of Fermat that every prime
number of the form 4n + 1 is a sum of two
squares.
Rashid Al Ghailani
Math 464 WI
Prof. Delaware
March 12th, 2024
1
Notes:
Leonhard Euler
1. When I recently 1 considered these numbers which arise from the
addition of two squares, I proved many properties which such numbers are
endowed with: it was not permitted to prolong adequately my thoughts about this
so that I could show for certain the truth of this theorem, which Fermat once
proposed to be proved via geometry. Nevertheless, I then published an attempt at
the proof, from which the certainty of this theorem shines much brighter, although
it is lacking, according to the criteria of a rigorous proof: I did not doubt that by
continuing in the same path, the desired proof could more easily be obtained,
which indeed since that time came to me with experience, so that the attempt, if
some other slick idea appears, may become a rigorous proof. Indeed, I can boast
of having provided nothing new about this subject, because Fermat himself
already claimed to have elicited a proof of this theorem, but in fact did not make it
public anywhere, just as with the many other exceptional discoveries of this man,
so that what now at last from these lost things we recover, as it were, these things
not unjustly are regarded as new discoveries. Because certainly no one ever so
successfully delved into the arcana of numbers as Fermat, all additional work in
this science to be developed appears to be spent in vain, except previously, which
things from this excellent man are now investigated, as if they are brought to light
anew. Although after him, many learned men in this field of studies have exerted
their own efforts, still generally nothing has followed which can be compared with
the ingenuity of this man.
2. As for the proof of the theorem, which I consider here, I have arranged
two propositions it is necessary to call upon for assistance, the proof of which I
have already given elsewhere. The first is that all numbers which are divisors of
sums of two squares prime between themselves are themselves sums of two
squares; thus, if a and b are numbers prime between themselves, and d is a divisor
of a number of the form aa+bb, then d will also be a sum of two squares; I have
2
given a proof of this theorem in a work mentioned earlier, in which I consider
numbers which are sums of two squares. The first proposition which the following
proof requires is that if p is a prime number and a and b are any numbers not
divisible by p, then ap−1 − bp−1 will always be divisible by the prime number p; I
have already given the proof of this result in Commentarii Academiae
Petropolitanae, Volume VIII.
3. So if 4n+1 is a prime number, all numbers contained in the form a4n−b4n
will be divisible by it, as long as neither of the numbers a or b is divisible by 4n +
1. Therefore, if a and b are numbers less than 4n + 1 but not equal to 0, then a
number of the form a4n−b4n will be divisible by the given prime number 4n + 1
without any limitation. Because a4n − b4n is a product of the factors a2n + b2n and a2n
− b2n, it is necessary that one of the two factors is divisible by 4n + 1; in other
words, it cannot be that neither or both have 4n + 1 as a divisor. For if indeed it
can be shown to be the case in which the form a2n+b2n is divisible by 4n+1, since
a2n +b2n, on account of its even exponent 2n, is a sum of two squares, neither of
which is divisible by 4n+1, from this it follows that the number 4n + 1 is a sum of
two squares.
4. Certainly the sum a2n + b2n will be divisible by 4n + 1 whenever the
difference a2n − b2n is not divisible by this same number. Therefore, anyone who
will say that the prime number 4n+1 is not a sum of two squares is forced to deny
that any number of the form a2n + b2n is divisible by 4n + 1. Thus, one should affirm
the same, namely, that all numbers contained in the form a2n−b2n are divisible by
4n+1, as long as neither a nor b is divisible by 4n+1. Accordingly, it is to be proved
here by me that not all numbers contained in the form a2n − b2n are divisible by 4n
+ 1; indeed, in this, if I will have succeeded. it will certainly be the case, if numbers
are substituted for a and b for which the form a2n − b2n is not divisible by 4n + 1;
then, in those cases, the other form a2n +b2n will necessarily be divisible by 4n+1.
From this, because a2n and b2n are squares, that which is proposed will have been
finished, namely, that the number 4n + 1 is a sum of two squares.
5. Therefore, so that I may prove that not all numbers contained in the form
a2n − b2n, in other words, not all differences between two powers with exponent
2n, are divisible by 4n + 1, I will consider the series of powers from
1 up to what is formed from the base 4
.
And now I declare that not all differences between two terms of this series are
divisible by 4n+1. For instance, if the terms of the first difference, 22n−1,32n−
were divisible by 4n + 1, then
the differences of the progression which are the second differences of that series
would also be divisible by 4n + 1; and by the same reasoning, third differences,
fourth, fifth, etc. would all be divisible by 4n + 1, and, finally, also differences of
order 2n, which, as they are constant, are all equal to each other. However, the
3
differences of order 2n are equal to 1 · 2 · 3 · 4···2n, which therefore are not
divisible by the prime number 4n + 1, from which it follows in turn that certainly
not all first differences are divisible by 4n + 1.
6. By virtue of which, the power of this proof is better observed, it is tobe
noted that the difference of order 2n is produced from 2n + 1 terms of the given
series, which, if they are taken from the beginning, are all so collected that
differences of whichever pairs should be divisible by 4n + 1 if the truth of the
theorem is negated. But if more terms from this last constituted difference were
assembled, and they were to proceed beyond the term (4n) 2 n, seeing that the
differences arising from the following term (4n + 1)2n do not relate to the
statement of the theorem, the proof would retain no validity. However, because
the last difference which we are considering depends only on 2n + 1 terms, the
conclusion which we deduced from it is entirely legitimate. And from this it
follows that the first differences will be like a2n−(a−1)2n, which is not divisible by
4n + 1, and in such a way that a is not greater than 2n + 1. However, later the sum
a2n + (a − 1)2n, is properly obtained, and therefore the sum of two squares is
necessarily divisible by 4n + 1, and thus the prime number 4n + 1 is a sum of two
squares.
7. Since the difference of order 2n depends on 2n + 1 terms from a series of
powers, let us consider only that many terms taken from the beginning of 1, 22n,
32n, 42n, 52n, 62n, 2n2n, (2n + 1)2n, from which the first differences will be 22n − 1, 32n
− 22n, 42n − 32n, 52n − 42n, …, (2n + 1)2n − (2n)2n, in which the number of terms in the
progression is 2n. And so, from the preceding demonstration it is evident that not
all terms in this progression of differences are divisible by the prime number 4n +
1. If however we examine the special cases in which 4n + 1 is a prime number, from
those differences, of which there are 2n, we will always discover that half are
divisible by 4n + 1, and the other half are, of course, not divisible. Even if this
observation does not seem to have the strength of a proof, it nevertheless does not
oppose at all what is to be explained. Therefore, it will be helpful to re-examine
some special cases.
8. The smallest prime number of the form 4n+1 is 5, which arises if n = 1.
From this will be obtained the two differences 2 2 − 1 and 32 − 22, of which the
former is not divisible by 5, but the other is so divisible. For the remaining cases,
I use d to indicate the differences which are divisible, 2 and 0 those which are not
divisible, the signs of which we write below the differences, according to the case.
4
From this it is evident that the divisible terms and the non-divisible terms are not restricted
by a fixed law, although the two are equal in multitude; nevertheless, it is evident that the
last term (2n+1)2n−(2n)2n is always divisible by 4n + 1, because it has the factor
(2n + 1)2 − 4nn = 4n + 1. But nothing certain can be determined about the others.
9. Moreover, as to the validity of this proof, it should be carefully observed
that that the proof to be examined is appropriate only if the number 4n + 1 is
prime, as the nature of the theorem certainly demands it. For if 4n + 1 were not
prime, it could not be asserted that it is a sum of two squares, nor that the form
a4n − b4n is necessarily divisible by it. Rather, the last conclusion which we
announced would be false, and those differences of order 2n, which are 1 · 2 · 3 ·
4···2n, are not divisible by 4n + 1. If however, 4n + 1 were not a prime number but
had factors which were less than 2n, then certainly the product 1 · 2 · 3 · 4···2n
would contain those factors and for that reason would be divisible by 4n + 1. But
if 4n + 1 is a prime number, then in fact one can assert that the product 1·2·3·4···2n
is clearly not divisible by 4n+1, because this product can be divided by no other
numbers unless they arise as factors in the product.
10. Finally, because this proof to be set forth rests on this foundation, that of
the series of powers 1,22n,32n,42n, etc., the differences of order 2n are constant and
all equal 1 · 2 · 3 · 4···2n; this can be seen more fully explained even if it is found
soundly exposited here and there in books of analysis. Therefore, first it is to be
noted that if the general term of any series or, in other words, that which
5
corresponds to an indefinite exponent of x, is Axm +Bxm−1 +Cxm−2 + Dxm−3 + Exm−4+
etc., then this series is said to be of degree m because m is the exponent of the
highest power of x. Then, if this general term is subtracted from the following A(x
+ 1)m + B(x + 1)m−1 + C(x + 1)m−2+ etc., the general term will produce a series of
differences in which the highest power of x will be m−1 and therefore the series
of differences will be of a lower degree, m−1. In a similar way, from the general
term of the series of first differences is collected a general term of the series of
second differences which, again, has an even lower degree, m − 2.
11. So if the given series is said to be of degree m, the series of first differences
will be of degree m−1, the series of second differences will in turn be of degree
m−2, the series of third differences of degree m−3, the series of fourth differences
will be of degree m − 4, and in general the series of nth order differences will be of
degree m − n. From this, the series of mth order differences will attain degree m−m
= 0, and therefore its general term, because the highest power of x is x0 = 1, will be
a constant quantity, and therefore all differences of order m will be equal to each
other. Hence, for series of first degree in which the general term is Ax+B, the first
differences will be equal to each other, while for series of second degree, which
are restricted to the general term Ax2 + Bx + C, the second differences are equal,
and so in turn.
12. If, therefore, we consider any series of powers.
, etc.
for which the general term is xm or, in other words, that which corresponds to a
base of x, the series of differences of order m will be constant from terms equal.
Now the general term of a series of first differences will be.
(x + 1)m − xm,
which, subtracted from the following
(x + 2)m − (x + 1)m
will give the general term of the series of second differences, which will be.
(x + 2)m − 2(x + 1)m + xm.
Thus, in turn, the general term of the series of third differences will be.
(x + 3)m − 3(x + 2)m + 3(x + 1)m − xm;
and finally, the general term of the series of differences of order m is.
6
+ etc.,
which, because they are constant quantities, will be the same for whatever number
is substituted for x; therefore, it will be either
+ etc.
or
+ etc.,
where in the former formula we set x = 0 and, in the latter, x = 1.
13. Let us now consider the special cases of this series and let us ascend
fromthe lowest powers to the highest. And first with m set equal to 1, the general
term of the first differences of the series 1, 2, 3, 4, 5, 6, etc. will be.
either
11 − 1 · 0 1 = 1
or
21 − 1 · 11 = 1.
If m = 2, the second differences of the series 1,22,32,42,52, etc. are
either
22 − 2 · 1 2
or
32 − 2 · 2 2 + 1 · 1 2 ;
but 22 − 2 · 12 = 2(21 − 1 · 11), from which these second differences are 2 · 1.
Let m = 3; the third differences of the series 1,23,33,43,53, etc. will be
either
33 − 3 · 2 3 + 3 · 1 3
or
43 − 3 · 3 3 + 3 · 2 3 − 1 · 1 3 ;
but
33 − 3 · 23 + 3 · 13 = 3(32 − 2 · 22 + 1 · 12) = 3 · 2 · 1,
because from the preceding case,
32 − 2 · 22 + 1 · 12 = 2 · 1.
In a similar fashion, if m = 4, the fourth differences of the series 1,24,34,44,54, etc.
will be
either
44 − 4 · 3 4 + 6 · 2 4 − 4 · 1 4
or
54 − 4 · 4 4 + 6 · 3 4 − 4 · 2 4 + 1 · 1 4 ;
but
44 − 4 · 34 + 6 · 24 − 4 · 14 = 4(43 − 3 · 33 + 3 · 23 − 1 · 13) = 4 · 3 · 2 · 1.
7
14. In order that this progression might be more easily observed, let the
differences of order m of the series 1,
, etc. equal P, and let the
m+1
m+1
differences of order m+1 of the series 1,2 ,3 ,4m+1,5m+1, etc. equal Q.
+ etc.,
,
where we expressed P in the former form and Q in the latter form. Here it is evident
that the number of terms is equal in each expression, and individual terms of the
expression P are to the individual terms of the expression Q as 1 is to m + 1. For
instance,
(m + 1)m : (m + 1)m+1 = 1 : m + 1
On account of this,
P_Q=1:m+1
and thus Q = (m + 1)P.
15. From this it is evident that
for the series
the differences
1, 2, 3, 4, 5, etc.
first = 1,
1,22,32,42,52, etc.
second = 1 · 2
1,23,33,43,53, etc.
third = 1 · 2 · 3
1,24,34,44,54, etc.
fourth= 1 · 2 · 3 · 4
, etc.
of order m = 1 · 2 · 3 · 4 · · · m
and therefore,
, etc.
of order 2n = 1 · 2 · 3 · 4 · · · 2n.
8
And thus we have also proved that the differences of order 2n in the series of
powers 1,22n,32n,42n,52n, etc. are not only constant but also equal to the product
1·2·3·4···2n, which we assumed in the proof of the proposed theorem. 3
Theorem 1
1. From the series of squares 1, 4, 9, 16, 25, etc., no numbers are divisible by
the prime number p, unless their roots are divisible by the same number p.
Proof
If for example some square number aa were divisible by the prime number p,
because it consists of the factors a and a, it would be necessary for one or the other
factor to be divisible by p. Therefore, the square number aa cannot be divisible by
the prime number p unless its root a is divisible by p.
Corollary 1
2. Thus, the square numbers arising from the roots p,2p,3p,4p, etc., namely,
pp,4pp,9pp,16pp, etc. are divisible by the prime number p, and all other square
numbers will not be divisible by the prime number p.
9
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