Problem:
Two positive integers differ by 60. The sum of their square roots is the square root of an integer that is not a perfect square. What is the maximum possible sum of the two integers?
Solution:
Let x represent the smaller of the two integers. Then x+x+60=y, and x+x+60+2x(x+60)=y. Thus x(x+60)=z2 for some positive integer z. It follows that
x2+60x=z2x2+60x+900=z2+900(x+30)2−z2=900, and (x+30+z)(x+30−z)=900
Thus (x+30+z) and (x+30−z) are factors of 900 with (x+30+z)>(x+30−z), and they are both even because their sum and product are even. Note that each pair of even factors of 900 can be found by doubling factor-pairs of 225 , so the possible values of (x+30+z,x+30−z) are (450,2),(150,6),(90,10), and (50,18). Each of these pairs yields a value for x which is 30 less than half their sum. These values are 196,48,20, and 4 . When x=196 or 4 , then x+x+60 is an integer. When x=48, we obtain 48+108=300, and when x=20, we obtain 20+80=180. Thus the desired maximum sum is 48+108=156.
OR
Let x represent the smaller of the two integers. Then x+x+60=y, and x+x+60+2x(x+60)=y. Thus x(x+60)=z2 for some positive integer z. Let d be the greatest common divisor of x and x+60. Then x=dm and x+60=dn, where m and n are relatively prime. Because dm⋅dn=z2, there are relatively prime positive integers p and q such that m=p2 and n=q2. Now d(q2−p2)=60. Note that p and q cannot both be odd, else q2−p2 would be divisible by 8 ; and they cannot both be even because they are relatively prime. Therefore p and q are of opposite parity, and q2−p2 is odd, which implies that q2−p2=1,3,5, or 15 . But q2−p2 cannot be 1 , and if q2−p2 were 15 , then d would be 4 , and x and x+60 would be squares. Thus q2−p2=3 or 5 , and (q+p,q−p)=(3,1) or (5,1), and then (q,p)=(2,1) or (3,2). This yields (x+60,x)=(22⋅20,12⋅20)=(80,20) or (x+60,x)=(32⋅12,22⋅12)=(108,48), so the requested maximum sum is 108+48=156.
OR
Let a and b represent the two integers, with a>b. Then a−b=60, and a+b=c, where c is an integer that is not a square. Dividing yields a−b=60/c. Adding these last two equations yields
2a=c+c60, so 2ac=c+60
Therefore ac is an integer, so c is even, as is ac, which implies ac is even. Hence c is a multiple of 4 , so there is a positive non-square integer d such that c=4d. Then
Thus d is a non-square divisor of 225 , so the possible values of d are 3,5,15, 45 , and 75. The maximum value of a, which occurs when d=3 or d=75, is 3+75+30=108, so the maximum value of b is 108−60=48, and the requested maximum sum is 48+108=156.