in how many ways 145² can be expressed as sum of squares of two natural numbers

You are looking for natural numbers a and b such that a²+b² = 145² which means (a,b,145) form a Pythagorean triplet.

Every Pythagorean triplet is of the form

Hence our task is here is to find out whether there exist numbers u,v such that u²+v² = 145 and u>v.

You only have to check for u = 12,11,10 and 9 (as 8² = 64< 1/2 * 145) and you can see that the only possible pairs are (12,1) and (9,8)

Thus there are two solutions   and   corresponding to these choices of u and v.

You only have to check for u = 12,11,10 and 9 (as 8² = 64< 1/2 * 145)???? Sir, will u PLZ explain this a bit more elaborately?

@hamba: Since a = u²-v²>0, we need u>v. So if u²+v² = 145, u²>145/2. Hence u>8. That is why checking for 12,11,10 and 9 suffices

@sourab: Well there just arent any other natural numbers. I guess they mean ordered pairs i.e they distinguish between (24,143) and (1433,24). Likewise (144,17) and (17,144) are distinctive pairs. That way its 4.

ok here are other two,



but how they got i dunno.

145²= (12²-1²)² + [2(12)(1)]²

and

145²= (9²-8²)² + [2(9)(8)]²

plz explain these

thnx

yeah, sachin alerted me about it and I remembered that the pair (u²-v², 2uv, u²+v²) is a primitive solution i.e. the gcd of the numbers is 1. So the general solution set is [k(u²-v²), 2kuv, k(u²+v²)]

So, we have to search for numbers k,u and v such that k(u²+v²) = 145

For k = 1 we have already obtained the solutions. Since k divides 145, k can now be only 5 or 29

If k = 5, then u²+v² = 29. So u = 5 and v = 3 is the only possibility and we obtain the solution (105,100)

Next for k = 29, u²+v² = 5 and for this we obtain a solution which sachin pointed out which is (116,87)

sorry, I somehow forgot that the solution set was for primitive roots for that equation. So the four pairs are

(105,100), (116,87),(143,24) and (144,17)

Thank u sir........