Can there be more than one Pythagorean triple such that the triple sum (a+b+c) is same?

If yes, please provide some examples.

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2 Answers

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Given x,y,a,bx,y,a,b such that x2+xy=a2+abx^2 + xy = a^2+ab, with x>yx > y and a>ba>b.

2(x2+xy)=2(a2+ab)⟹(x2+y2)+2xy+(x2−y2)=(a2+b2)+2ab+(a2−b2)2(x^2+xy) = 2(a^2+ab) \implies (x^2+y^2) + 2xy + (x^2-y^2) = (a^2+b^2) + 2ab + (a^2-b^2). The three terms on each side form a triple.

For example:

Let x=8,y=7,a=10,b=2x=8,y=7,a=10,b=2. Then, 113+112+15=104+40+96113+112 + 15 = 104+40+96. Furthermore, 152+1122=113215^2 + 112^2 = 113^2 and 402+962=104240^2+96^2=104^2.

More exciting: Let x=48,y=44,a=64,b=5x=48,y=44,a=64,b=5. Then, 4224+368+4240=640+4071+41214224+ 368 + 4240 = 640+4071+4121. Further 42242+3682=424024224^2 + 368^2 = 4240^2 and 6402+40712=41212640^2+4071^2=4121^2.

Even bigger: Let x=87,y=43,a=78,b=67x=87,y=43,a=78,b=67. Then, 7482+5720+9418=10452+1595+105737482+ 5720 + 9418 = 10452+1595+10573. Further 74822+57202=941827482^2 + 5720^2 = 9418^2 and 104522+15952=10573210452^2+1595^2=10573^2.

Finally, the biggest: x=99,y=61,a=96,b=69x=99,y=61,a=96,b=69. Then, 12078+6080+13522=13248+4455+1397712078+ 6080 + 13522 = 13248+4455+13977. Further 120782+60802=13522212078^2 + 6080^2 = 13522^2 and 132482+44552=13977213248^2+4455^2=13977^2.

You can explore further.

EDIT : Just adding another : x=10000,y=287,a=10125,b=35x=10000 ,y= 287 ,a=10125 ,b= 35 , with 5740000+99917631+100082369=708750+102514400+1025168505740000 + 99917631+100082369=708750+102514400+102516850.

Let a=r2−s2a = r^2 – s^2, b=2rsb = 2rs and c=r2+s2c = r^2 + s^2 be a triplet whose sum is equal to that of another triplet l=x2−y2l = x^2 – y^2, m=2xym = 2xy and n=x2+y2n = x^2 + y^2. Then a+b+c=l+m+na+b+c = l+m+n; which gives

r(r+s)=x(x+y)

r(r+s) = x(x+y)

The above equation is nothing but representing a number as the product of two factors in two different ways.

Thus if you take any number that has four or more divisors which can be represented as above then you can construct two Pythagorean triplet which have the same sum by plugging the values r,s,lr, s, l and mm in the above formula for a,b,c,l,ma,b,c,l,m and nn.