* a Ms se * 
y' 
Solution of a Diophantine Problem. 295 
we 
Arr. XI.—An easy solution of a Diophantine Problem; by A.D. 
Wueeter, Principal of the Latin Grammar School, Salem, Mass. _ 
Problem.—To find two squares, whose sum shall be a square; or 
in other words, to find rational values for the legs and hypothenuse 
of a right angled triangle. ie ns 
Rule—Take any two numbers, of which the difference is 2. 
Their sum will be the root of one square; their product, that of the 
other. Add 2 to the product, just found, and you obtain the root of 
the sum of the squares, or the value of the hypothenuse. . 
Example.—Take the numbers 10 and 12; then 10-+-12=22= 
the root of one square, and 10X12=120= that ‘of the other. 
Furthermore, 120-+2=122= the root of their sum; since (22)* 4 
+(120)?=(122)2. These quantities may be multiplied by any — 
to the conditions of the problem. ‘i 
Demonstration.—Let z represent the smaller number, and z+2, 
the larger, Their sum z+ (z-+2)=22+2 [A], and their product, 
2.(z+2)=224922[B]. Now (2z+2)?+(z?+2z)?=2'4 423-4 
827+8z44= a square, whose root is z?-+2z-+2=(z?+2z)4+ 
2[C] = the product [B]+2. Multiplying the expressions [A], 
[B], and [C], by m, and squaring, we have m?(2z+4-2)? +-m?(z? + 
22)? =m? (4414294822 +8244) =m? (z2+22+2)*, whatever 
be the values of m and z. Q. E. D. 
When m=0, and z is an odd number, the quantities [A], [B], . 
[C], are prime to each other. But when 2 is even, these quantities 
may be divided by 2, and by no other number. For, if we suppose 
the quantity [B] to be divisible by n, then, when nis a prime number, 
either 2 or +2 must be divisible by it. , Because, if a prime num- 
ber will divide neither factor, it cannot divide the product. (Euler, - 
App. 10). 
ate 
“ But since the difference between z and z+2 is only 2, 
ii rn, when greater than 2, cannot divide them both; and, 
~ fonseq ntly, cansioi divide the quantity [A]. (Bonnycastle, p. 145). 
Again if [B] and [C] are divisible by , then the parts of [C], (27+ 
22) and 2 may be divided by n. But, it is plain, that this cannot be 
done,, unle s, as before, n=2. 
I n a prime number, it is necessary to remark, that all 
F Compound numbers may be resolved into prime factors, each of 
; 
Bs 48 e fi. 
° i ° . ; 
number whatever, and their products, when squared, will still answer x 
