Le 4 mai 2021, la plateforme Yahoo Questions/Réponses fermera. Elle est désormais accessible en mode lecture seule. Aucune modification ne sera apportée aux autres sites ou services Yahoo, ni à votre compte Yahoo. Vous trouverez plus d’informations sur l'arrêt de Yahoo Questions/Réponses et sur le téléchargement de vos données sur cette page d'aide.
Like number theory problems?
Let A be an even integer such that A^2 + 1 be composite. Can you always find (a,b) integers with b odd, such that a^2+b^2 = A^2+1, and 10*b > A.
Related to
http://answers.yahoo.com/question/index;_ylt=AqKV8...
If not, counterexamples welcome, smallest wins...
@ Josh K. The question is for all A such that xxx, consider the highest b: do you necessarily have b/A > 1/10.
So either you prove it for all A, not just A = 8, or you disprove it, which Rita the Dog did.
2 réponses
- Rita the dogLv 7il y a 8 ansRéponse favorite
If I understand correctly, then A=1130 is a counter example. A^2+1 = 1276901 = 577*2213, composite. The only decompositions of A^2+1 into a sum of two squares are 95^2+1126^2 and 1^2+1130^2 so the best b is b=95 and 10*b=950<1130=A.
Unless I am off track, each of the following values of A is also a counterexample:
1130, 1624, 2284, 2630, 3296, 3486, 4036, 4640, 5050, 5746, 6376, 6706, 6744, 7124, 7746, 8036, 8040, 8950, 9604
A more extreme counterexample is A=83030, where largest b=815, so 100*b<A.
- ?Lv 6il y a 8 ans
a^2+b^2 = A^2+1
so doesn't b = (+/-) 1
Am I missing something to your question?
Then let a = 0, a^2+b^2 = 1 not composite
a = 2, a^2 + b^2 = 5, not composite
a = 4, a^2 + b^2 = 17 not composite
a = 6, a^2 + b^2 = 37, not composite
a = 8, a^2 + b^2 = 65 = 13*5 composite
a = 8
10*1 > 8
So If I am reading this problem correctly, here is a solution.
But the question was can you ALWAYS.
I have found counterexamples, so the universal is disproven.
But I have also shown existance.
