Every perfect square is the sum of two squares: x2 =x2+02. Indeed, this was one of the “base cases” that we used in proving Theorem 9.1 determining those positive integers which are sums of two squares. But things are very different if we ask which perfect squares are the sum of two squares of positive integers. The smallest such is 25: 32+42=52.
The equationx2+y2=z2is associated with Pythagoras. Not only did he prove his famous theorem asserting that this holds if z is the hypotenuse of a right- angled triangle and x and y are the other two sides, but he also gave a rule for finding all the solutions of this equation in positive integers. (The famous solution 32+42=52gives a right-angled triangle which had been used by surveyors since before the time of Pythagoras. Take a loop of string with twelve equally-spaced knots. Taking hold of the appropriate knots and pulling the string tight gives a right angle.)
Theorem 9.4 Let x, y and z be positive integers satisfying x2+y2=z2. Then there are positive integers d,s,t withgcd(s,t) =1, such that, after interchanging x and y if necessary, we have
x=2std, y= (s2−t2)d, z= (s2+t2)d. Note thats=2,t=1,d=1 gives the solution(x,y,z) = (3,4,5).
Proof We use the following principle: if ab=c2 and gcd(a,b) =1, then each ofaandbis a square. For any prime divisor ofaboccurs to an even power, and must occur in one ofaandband not the other; so each ofaandbis a product of even powers of primes, and so is a square. More generally, if the product of any number of pairwise coprime factors is a square, then each factor is a square.
9.6. PYTHAGORAS AND FERMAT 97 In a similar way, ifab=c2where gcd(a,b) =2, thena=2m2andb=2n2for some integers m,n. (Just replaceaandb bya/2 andb/2; these are coprime and their product is(c/2)2, so each is a square.)
Now suppose thatx2+y2=z2.
• any common factor of two ofx,y,zmust divide the third, and we can divide through by it and get a smaller solution. So we can assume thatx,y,z are pairwise coprime, by dividing by their gcd (sayd).
• Since squares are congruent to 0 or 1 mod 4, the only solutions to our equa- tion mod 4 are 0+0=0 and 0+1=1. Since the variables are pairwise coprime, we can assume the latter: that is, (swappingxandyif necessary) xis even,yandzare odd.
• Nowx2=z2−y2= (z+y)(z−y), and gcd(z+y)(z−y) =2. (For bothz+y andz−yare even, so there is a common factor 2; and ifdis the gcd, thend divides both(z+y) + (z−y) =2zand(z+y)−(z−y) =2y, sod=2.) So z−y=2s2andz+y=2t2for some (coprime) integerssandt. Now
• ◦ x2=4s2t2, sox=2st;
◦ y= ((z+y)−(z−y))/2=s2−t2;
◦ z= ((z+y) + (z−y))/2=s2+t2. In the seventeenth century, Pierre de Fermat wrote a note in the margin of his copy of the book on number theory by the Greek mathematician Diophantus. The note was opposite the place where Diophantus gave the preceding theorem of Pythagoras. Fermat claimed that he had a “truly wonderful” proof that, for any n>2, the equationxn+yn=znhas no solution in positive integers, but the margin where he was writing was too small to contain it.
Mathematicians took up the challenge of trying to find the proof of what be- came known, ironically, as Fermat’s Last Theorem. Finally in the 1990s, Andrew Wiles succeeded in finding a proof. But his proof was very long and compli- cated, and used many concepts which had not been invented in Fermat’s time. Moreover, no evidence of such a proof was ever found in Fermat’s papers. It is generally believed now that he didn’t have a proof; perhaps he thought he had one but it contained a mistake.
We certainly cannot prove Wiles’ Theorem here. But as an illustration, we prove a simple case:
98 CHAPTER 9. SUMS OF SQUARES
Proof We actually consider a slightly different equation, namelyx4+y4=z2. If we show that this equation has no solution, then the equation of the theorem has no solution either, since if x,y,zsatisfy the equation in the theorem, then x,y,z2 satisfy the modified equation.
Suppose that x4+y4=z2, where x,y,z are positive integers. We may sup- pose that this is the solution with the smallest possible value of z. Then x,y,z are pairwise coprime, since if there were a common prime factor p of any two of them it would divide the third and we could replace the solution (x,y,z) by (x/p,y/p,z/p2). So one ofx2andy2is even; without loss of generality,xis even.
Since(x2)2+ (y2)2=z2, we can apply Pythagoras to conclude that x2=2st, y2=s2−t2, z=s2+t2,
where gcd(s,t) =1.
Applying Pythagoras to the equationt2+y2=s2(remembering thatyis odd), we have
t=2uv, y=u2−v2, s=u2+v2,
where gcd(u,v) =1. It follows that gcd(u,u2+v2) =gcd(v,u2+v2) =1. Then x2=2st =4uv(u2+v2), so that uv(u2+v2) is a square. Since the factors are pairwise coprime, we haveu=m2,v=n2, andu2+v2=r2. Thus
m4+n4=r2.
But r ≤u2+v2= s<s2+t2 =z, so we have (m,n,r) is a solution of the original equation smaller than the solution (x,y,z), which we assumed to be the smallest. This contradiction shows that no solution can exist.